神级代码推荐

这些是我看到的一些有趣的代码

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*/
#include 
using namespace std;

#define INF        (~(1<<31))
#define INFLL      (~(1ll<<63))
#define pb         push_back
#define mp         make_pair
#define abs(a)     ((a)>0?(a):-(a))
#define lalal      puts("*******");
#define s1(x)      scanf("%d",&x)
#define Rep(a,b,c) for(int a=(b);a<=(c);a++)
#define Per(a,b,c) for(int a=(b);a>=(c);a--)
#define pair1      first
#define pair2      second

typedef long long int LL ;
typedef unsigned long long int uLL ;

const LL     MOD = 1e18;
const int    N   = 100000+5;
const double eps = 1e-8;
const double PI  = acos(-1.0);

inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void fre(){
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
}
inline LL _gcd(LL a,LL b){return (b==0)?a:_gcd(b,a%b);}
inline LL _lcm(LL a,LL b){return (a/_gcd(a,b)*b);    }
/***********************************************************************/
struct Complex{
    double real, image;
    Complex(double _real, double _image){
        real = _real;
        image = _image;
    }
    Complex(){}
};

Complex *AA = new Complex[N<<2],*BB = new Complex[N<<2];
Complex operator + (const Complex &c1, const Complex &c2){
    return Complex(c1.real + c2.real, c1.image + c2.image);
}
Complex operator - (const Complex &c1, const Complex &c2){
    return Complex(c1.real - c2.real, c1.image - c2.image);
}
Complex operator * (const Complex &c1, const Complex &c2){
    return Complex(c1.real*c2.real - c1.image*c2.image, c1.real*c2.image + c1.image*c2.real);
}

int rev(int id, int len){
    int ret = 0;
    for(int i = 0; (1 << i) < len; i++){
        ret <<= 1;
        if(id & (1 << i)) ret |= 1;
    }
    return ret;
}
Complex A[N<<2];
void FFT(Complex *a, int len, int DFT)
{
    for(int i = 0; i < len; i++)
        A[rev(i, len)] = a[i];
    for(int s = 1; (1 << s) <= len; s++)
    {
        int m = (1 << s);
        Complex wm = Complex(cos(DFT*2*PI/m), sin(DFT*2*PI/m));
        for(int k = 0; k < len; k += m)
        {
            Complex w = Complex(1, 0);
            for(int j = 0; j < (m >> 1); j++)
            {
                Complex t = w*A[k + j + (m >> 1)];
                Complex u = A[k + j];
                A[k + j] = u + t;
                A[k + j + (m >> 1)] = u - t;
                w = w*wm;
            }
        }
    }
    if(DFT == -1) for(int i = 0; i < len; i++) A[i].real /= len, A[i].image /= len;
    for(int i = 0; i < len; i++) a[i] = A[i];
    return;
}

int a[N],b[N];
int ans[N<<2];

int main(){
    int n,m;
    while(~scanf("%d %d",&n,&m)){
        int la = n+1,lb = m+1;
        int sa,sb;
        sa=sb=0;
        while((1<while((1<int len = (1<<(max(sa,sb)+1));

        for(int i=0;i"%d",&a[i]);
        for(int i=0;i"%d",&b[i]);

        for(int i=0;i1]):0),0);
            BB[i]=Complex(((i1]):0),0);
        }

        FFT(AA,len,1);
        FFT(BB,len,1);
        for(int i=0;i*BB[i],ans[i]=0;
        FFT(AA,len,-1);
        for(int i=0;iint)(AA[i].real+0.5);

        for(int i=la+lb-2;i>=0;--i)  printf("%d%c",ans[i],i?' ':'\n');
    }
    return 0;
}
/**
1 1
3 1
1 9

*/
/*
                   _ooOoo_
                  o8888888o
                  88" . "88
                  (| -_- |)
                  O\  =  /O
               ____/`---'\____
             .'  \\|     |//  `.
            /  \\|||  :  |||//  \
           /  _||||| -:- |||||-  \
           |   | \\\  -  /// |   |
           | \_|  ''\---/''  |   |
           \  .-\__  `-`  ___/-. /
         ___`. .'  /--.--\  `. . __
      ."" '<  `.___\_<|>_/___.'  >'"".
     | | :  `- \`.;`\ _ /`;.`/ - ` : | |
     \  \ `-.   \_ __\ /__ _/   .-` /  /
======`-.____`-.___\_____/___.-`____.-'======
                   `=---='
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
         佛祖保佑       永无BUG
*/

排版好差
自己粘下来看吧

/*  Number theory template :
    Deals with various aspects of integer, division, modulo, etc.
*/

#include 

namespace number {

    const double PI = acos (-1);

    /*  Basic constants & functions :
            long long inverse (const long long &x, const long long &mod) :
                returns the inverse of x modulo mod.
                i.e. x * inv (x) % mod = 1.
            int fpm (int x, int n, int mod) :
                returns x^n % mod. i.e. Fast Power with Modulo.
            void euclid (const long long &a, const long long &b,
                         long long &x, long long &y) :
                solves for ax + by = gcd (a, b).
            long long gcd (const long long &a, const long long &b) :
                solves for the greatest common divisor of a and b.
            long long mul_mod (const long long &a, const long long &b, const long long &mod) :
                returns a * b % mod.
            long long llfpm (const long long &x, const long long &n, const long long &mod) :
                returns x^n % mod.
    */

    long long abs (const long long &x) { return x > 0 ? x : -x; }

    long long inverse (const long long &x, const long long &mod) {
        if (x == 1) return 1;
        return (mod - mod / x) * inverse (mod % x, mod) % mod;
    }

    int fpm (int x, int n, int mod) {
        register int ans = 1, mul = x;
        while (n) {
            if (n & 1) ans = int (1ll * ans * mul % mod);
            mul = int (1ll * mul * mul % mod);
            n >>= 1;
        }
        return ans;
    }

    void euclid (const long long &a, const long long &b,
                 long long &x, long long &y) {
        if (b == 0) x = 1, y = 0;
        else euclid (b, a % b, y, x), y -= a / b * x;
    }

    long long gcd (const long long &a, const long long &b) {
        if (!b) return a;
        long long x = a, y = b;
        while (x > y ? (x = x % y) : (y = y % x));
        return x + y;
    }

    long long mul_mod (const long long &a, const long long &b, const long long &mod) {
        long long ans = 0, add = a, k = b;
        while (k) {
            if (k & 1) ans = (ans + add) % mod;
            add = (add + add) % mod;
            k >>= 1;
        }
        return ans;
    }

    long long llfpm (const long long &x, const long long &n, const long long &mod) {
        long long ans = 1, mul = x, k = n;
        while (k) {
            if (k & 1) ans = mul_mod (ans, mul, mod);
            mul = mul_mod (mul, mul, mod);
            k >>= 1;
        }
        return ans;
    }

    /*  Discrete Fourier transform :
            int dft::init (int n) : initializes the transformation with dimension n.
            void dft::solve (complex *a, int n, int f) :
                transforms array a with dimension n to its image representation.
                Transforms back when f = 1.
    */

    template <int MAXN = 1000000>
    struct dft {

        typedef std::complex <double> complex;

        static complex e[2][MAXN];

        int init (int n) {
            int len = 1;
            for (; len <= 2 * n; len <<= 1);
            for (int i = 0; i < len; i++) {
                e[0][i] = complex (cos (2 * PI * i / len), sin (2 * PI * i / len));
                e[1][i] = complex (cos (2 * PI * i / len), -sin (2 * PI * i / len));
            }
            return len;
        }

        void solve (complex *a, int n, int f) {
            for (int i = 0, j = 0; i < n; i++) {
                if (i > j) std::swap (a[i], a[j]);
                for (int t = n >> 1; (j ^= t) < t; t >>= 1);
            }
            for (int i = 2; i <= n; i <<= 1)
                for (int j = 0; j < n; j += i)
                    for (int k = 0; k < (i >> 1); k++) {
                        complex A = a[j + k];
                        complex B = e[f][n / i * k] * a[j + k + (i >> 1)];
                        a[j + k] = A + B;
                        a[j + k + (i >> 1)] = A - B;
                    }
            if (f == 1) {
                for (int i = 0; i < n; i++)
                    a[i] = complex (a[i].real () / n, a[i].imag ());
            }
        }
    };

    /* Number-theoretic transform :
        void ntt::solve (int *a, int n, int f, int mod, int prt) :
            transfers a[n] to its image representation.
            Converts back if f = 1.
            Requries specific mod and corresponding prt to work. (given in MOD and PRT)
        int ntt::crt (int *a, int mod) :
            fixes the results a from module 3 primes to a certain module mod.
    */

    template <int MAXN = 1000000>
    struct ntt {

        void solve (int *a, int n, int f, int mod, int prt) {
            for (register int i = 0, j = 0; i < n; i++) {
                if (i > j) std::swap (a[i], a[j]);
                for (register int t = n >> 1; (j ^= t) < t; t >>= 1);
            }
            for (register int i = 2; i <= n; i <<= 1) {
                static int exp[MAXN];
                exp[0] = 1;
                exp[1] = fpm (prt, (mod - 1) / i, mod);
                if (f == 1) exp[1] = fpm (exp[1], mod - 2, mod);
                for (register int k = 2; k < (i >> 1); k++) {
                    exp[k] = int (1ll * exp[k - 1] * exp[1] % mod);
                }
                for (register int j = 0; j < n; j += i) {
                    for (register int k = 0; k < (i >> 1); k++) {
                        register int &pA = a[j + k], &pB = a[j + k + (i >> 1)];
                        register int A = pA, B = int (1ll * pB * exp[k] % mod);
                        pA = (A + B) % mod;
                        pB = (A - B + mod) % mod;
                    }
                }
            }
            if (f == 1) {
                register int rev = fpm (n, mod - 2, mod);
                for (register int i = 0; i < n; i++) {
                    a[i] = int (1ll * a[i] * rev % mod);
                }
            }
        }

        int MOD[3] = {1045430273, 1051721729, 1053818881}, PRT[3] = {3, 6, 7};

        int crt (int *a, int mod) {
            static int inv[3][3];
            for (int i = 0; i < 3; i++)
                for (int j = 0; j < 3; j++)
                    inv[i][j] = (int) inverse (MOD[i], MOD[j]);
            static int x[3];
            for (int i = 0; i < 3; i++) {
                x[i] = a[i];
                for (int j = 0; j < i; j++) {
                    int t = (x[i] - x[j] + MOD[i]) % MOD[i];
                    if (t < 0) t += MOD[i];
                    x[i] = int (1LL * t * inv[j][i] % MOD[i]);
                }
            }
            int sum = 1, ret = x[0] % mod;
            for (int i = 1; i < 3; i ++) {
                sum = int (1LL * sum * MOD[i - 1] % mod);
                ret += int (1LL * x[i] * sum % mod);
                if (ret >= mod) ret -= mod;
            }
            return ret;
        }

    };

    /*  Chinese remainder theroem :
            bool crt::solve (const std::vector  > &input,
                            std::pair &output) :
                solves for an integer set x = output.first + k * output.second
                    that satisfies x % input[i].second = input[i].first.
                Returns whether a solution exists.
    */

    struct crt {

        long long fix (const long long &a, const long long &b) {
            return (a % b + b) % b;
        }

        bool solve (const std::vector <std::pair <long long, long long> > &input,
                    std::pair <long long, long long> &output) {
            output = std::make_pair (1, 1);
            for (int i = 0; i < (int) input.size (); ++i) {
                long long number, useless;
                euclid (output.second, input[i].second, number, useless);
                long long divisor = gcd (output.second, input[i].second);
                if ((input[i].first - output.first) % divisor) {
                    return false;
                }
                number *= (input[i].first - output.first) / divisor;
                number = fix (number, input[i].second);
                output.first += output.second * number;
                output.second *= input[i].second / divisor;
                output.first = fix (output.first, output.second);
            }
            return true;
        }

    };

    /*  Miller Rabin :
            bool miller_rabin::solve (const long long &) :
                tests whether a certain integer is prime.
    */

    struct miller_rabin {

        int BASE[12] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};

        bool check (const long long &prime, const long long &base) {
            long long number = prime - 1;
            for (; ~number & 1; number >>= 1);
            long long result = llfpm (base, number, prime);
            for (; number != prime - 1 && result != 1 && result != prime - 1; number <<= 1)
                result = mul_mod (result, result, prime);
            return result == prime - 1 || (number & 1) == 1;
        }

        bool solve (const long long &number) {
            if (number < 2) return false;
            if (number < 4) return true;
            if (~number & 1) return false;
            for (int i = 0; i < 12 && BASE[i] < number; ++i)
                if (!check (number, BASE[i]))
                    return false;
            return true;
        }

    };

    /*  Pollard Rho :
            std::vector  pollard_rho::solve (const long long &) :
                factorizes an integer.
    */

    struct pollard_rho {

        miller_rabin is_prime;
        const long long threshold = 13E9;

        long long factorize (const long long &number, const long long &seed) {
            long long x = rand() % (number - 1) + 1, y = x;
            for (int head = 1, tail = 2; ; ) {
                x = mul_mod (x, x, number);
                x = (x + seed) % number;
                if (x == y)
                    return number;
                long long answer = gcd (abs (x - y), number);
                if (answer > 1 && answer < number)
                    return answer;
                if (++head == tail) {
                    y = x;
                    tail <<= 1;
                }
            }
        }

        void search (const long long &number, std::vector<long long> &divisor) {
            if (number > 1)  {
                if (is_prime.solve (number))
                    divisor.push_back (number);
                else {
                    long long factor = number;
                    for (; factor >= number;
                            factor = factorize (number, rand () % (number - 1) + 1));
                    search (number / factor, divisor);
                    search (factor, divisor);
                }
            }
        }

        std::vector <long long> solve (const long long &number) {
            std::vector <long long> ans;
            if (number > threshold)
                search (number, ans);
            else {
                long long rem = number;
                for (long long i = 2; i * i <= rem; ++i)
                    if (number % i)
                        ++i;
                    else {
                        ans.push_back (i);
                        rem %= i;
                    }
            }
            return ans;
        }

    };

    /*  Adaptive Simpson's method :
            double simpson::solve (double (*f) (double), double l, double r, double eps) :
                integrates f over (l, r) with error eps.
    */

    struct simpson {

        double area (double (*f) (double), double l, double r) { 
            double m = l + (r - l) / 2;
            return (f (l) + 4 * f (m) + f (r)) * (r - l) / 6;
        }

        double solve (double (*f) (double), double l, double r, double eps, double a) {
            double m = l + (r - l) / 2;
            double left = area (f, l, m), right = area (f, m, r);
            if (fabs (left + right - a) <= 15 * eps) return left + right + (left + right - a) / 15.0;
            return solve (f, l, m, eps / 2, left) + solve (f, m, r, eps / 2, right);
        }

        double solve (double (*f) (double), double l, double r, double eps) {
            return solve (f, l, r, eps, area (f, l, r));
        }

    };

}

using namespace number;

typedef std::complex <double> complex;
template <int MAXN> std::complex <double> dft ::e[2][MAXN];
const int MAXN = 1000000;

dft <> m_dft;
int N, M;
complex A[MAXN], B[MAXN], C[MAXN];

int main () {
    scanf ("%d%d", &N, &M);
    for (int i = 0; i < N + 1; ++i) {
        int tmp;
        scanf ("%d", &tmp);
        A[i] = complex (tmp, 0);
    }
    for (int i = 0; i < M + 1; ++i) {
        int tmp;
        scanf ("%d", &tmp);
        B[i] = complex (tmp, 0);
    }
    int size = N + M + 1;
    int len = m_dft.init (size);
    m_dft.solve (A, len, 0);
    m_dft.solve (B, len, 0);
    for (int i = 0; i < len; ++i)
        C[i] = A[i] * B[i];
    m_dft.solve (C, len, 1);
    for (int i = 0; i < N + M + 1; ++i)
        printf ("%d%c", int (C[i].real () + .5), " \n"[i == size - 1]);
    return 0;
}

超级强的模板

#include 
#include 
#include 
#include 
#define st(now) ((isvar[now])?var[st[now]]:st[now])
using namespace std;
int In,input[1000],len,dep=0,funsum,retu,PA,read=0;bool RETU;
int varsum[100000],var[3000000],funstart[1000],pa[1000],pasum[1000],wei[3000000][20],weisum[3000000];
string par[1000][100],code;
set<char> s1;//实词集合（控制、函数、变量） 
set<char> s2;//表达式字符集合 
set<char> s3;//数字集合 
map<string,int> m1[100000];//变量映射 
map<string,int> m2;//函数映射 
void init()//初始化 
{
    In=2;
    cin >> input[1] >> input[2];
    code="int main(){ int a,b;cin >> a >> b;cout << a+b;}";
    bool flag=1;
    for(char ch='a';ch<='z';ch++) s1.insert(ch),s2.insert(ch);
    for(char ch='A';ch<='Z';ch++) s1.insert(ch),s2.insert(ch);
    for(char ch='0';ch<='9';ch++) s1.insert(ch),s2.insert(ch),s3.insert(ch);
    s1.insert('_');s2.insert('_');
    s2.insert('(');s2.insert(')');
    s2.insert('+');s2.insert('-');s2.insert('*');s2.insert('/');s2.insert('%');
    s2.insert('!');s2.insert('&');s2.insert('|');s2.insert('=');s2.insert('<');s2.insert('>');s2.insert('^');
    s2.insert('[');s2.insert(']');
    code.erase(0,1);
    for(int i=0;i<=code.length();)
        if((code[i]==' '||code[i]=='\n'||code[i]=='\t')&&((!s1.count(code[i-1])) || (!s1.count(code[i+1]))))
            code.erase(i,1);
        else i++;
    varsum[0]=1;len=code.length();RETU=0;PA=0;
}
int level(int &now)//符号转成运算级 
{
    switch(code[now])
    {
        case'<':if(code[++now]=='=')return 6;
                    else {now--;return 8;}
        case'>':if(code[++now]=='=')return 7;
                    else {now--;return 9;}
        case'=':if(code[++now]=='=')return 4;
                    else {now--;return 0;}
        case'!':now++;return 5;
        case'|':now++;return 1;
        case'&':now++;return 2;
        case'^':return 3;
        case'+':return 10;
        case'-':return 11;
        case'*':return 12;
        case'/':return 13;
        case'%':return 14;
    }
}
bool dayu(int a,int b)
{
    if(a==0) return 0;
    if(a>=b) return 1;
    if((a==4)&&(b==5))return 1;
    if((a>=6)&&(a<=9)&&(b>=6)&&(b<=9)) return 1;
    if((a==10)&&(b==11)) return 1;
    if((a>=12)&&(a<=14)&&(b>=12)&&(b<=14)) return 1;
    return 0;
}
int word(int now)//找下一个单词（变量 int 函数之类的） 
{
    int end;
    for(end=now;s1.count(code[end]);end++);
    return end;
}
int express(int now)//找下一个表达式 
{
    int end;
    for(end=now;s2.count(code[end]);end++);
    return end;
}
int pass(int now)
{
    while(code[now]!=';') now++;
    return now+1;
}
int block(int now)//跳过一段程序 
{
    if(code[now]=='{')
    {
        int dep=1;
        for(now++;dep;now++) if(code[now]=='{') dep++;else if(code[now]=='}') dep--;
        return now;
    }
    else
    {
        int next=word(now);
        if((next-now==3 && code.substr(now,3)=="for")||(next-now==5 && code.substr(now,5)=="while"))
        {
            int p=next+1,deep=1;
            while(deep)
            {
                if(code[p]=='(') deep++;
                if(code[p]==')') deep--;
                p++;
            }
            return block(p);
        }
        else
        if(next-now==2 && code.substr(now,2)=="if")
        {
            int p=next+1,deep=1;
            while(deep)
            {
                if(code[p]=='(') deep++;
                if(code[p]==')') deep--;
                p++;
            }
            p=block(p);
            next=word(p);
            if(next-p==4 && code.substr(p,4)=="else")
                return block(next);
            else return p;
        }
        else
        {
            int p=now;
            while(code[p]!=';') p++;
            return p+1;
        }
    }
}
int num(int &now)//得到最近的十进制数 
{
    int sum=0;
    while(s3.count(code[now]))
        sum=sum*10+code[now++]-'0';
    return sum;
}
int getsize(int &now,int k)//得到数组的大小 
{
    int sum=1;
    while(code[now]=='[')
    {
        now++;
        int tem=num(now);
        wei[k][++weisum[k]]=tem;
        sum*=tem;
        now++;
    }
    return sum;
}
int focus(int &next,int first)//得到所要的变量在数组中的位置 
{
    int calc(int &now);
    int sum=0,SUM=1,i=0;
    if(code[next]=='[')
    {
        while(code[next]=='[')
        {
            next++;i++;
            sum+=SUM*calc(next);
            SUM*=wei[first][i];
            next++;
        }
    }
    return sum;
}
int newfun(int beg,int end)//定义新函数 
{
    m2[code.substr(beg,end-beg)]=++funsum;int i=0;
    for(int next=word(end+1);code[end++]!=')';end=next,next=word(end+1))
        if(next-end!=3 || code.substr(end,3)!="int")
            par[funsum][++i]=code.substr(end,next-end); 
    pasum[funsum]=i;
    funstart[funsum]=end++;
    for(int dep=1;dep;end++) if(code[end]=='{') dep++;else if(code[end]=='}') dep--;
    return end;
}
int newvar(int beg)//定义新变量 
{
    int end;
    for(end=beg;code[end]!=';';beg=end+1)
    {
        end=word(beg),m1[dep][code.substr(beg,end-beg)]=varsum[dep];
        int from=varsum[dep];weisum[varsum[dep]]=0;
        for(varsum[dep]+=getsize(end,varsum[dep]);from<=varsum[dep];from++) var[from]=0;
    }
    return end+1;
}
int newdep()//新的一层 
{
    varsum[dep+1]=varsum[dep];m1[dep+1]=m1[dep];dep++;
}
int calc(int &now)//表达式求值 
{
    int run(int now,bool isfun);
    int fh[20],st[20];bool isvar[20];int top=0,pre[10],presum=0;
    for(;code[now]!=';' && code[now]!=')' &&(code[now]!='<' || code[now+1]!='<')&&(code[now]!=']')&&(code[now]!=',');)
        if(s3.count(code[now]))
        {
            st[++top]=num(now);
            while(presum>0)
            {
                if(pre[presum]==1) st[top]=!st[top];
                if(pre[presum]==2) st[top]=-st[top];
                presum--;
            }
            isvar[top]=0;fh[top]=-1;
        }
        else
        if(s1.count(code[now]))
        if(m1[dep].count(code.substr(now,word(now)-now)))
        {
            int next=word(now),first=m1[dep][code.substr(now,word(now)-now)];
            st[++top]=first+focus(next,first);isvar[top]=1;
            while(presum>0)
            {
                st[top]=st(top);isvar[top]=0;
                if(pre[presum]==1) st[top]=!st[top];
                if(pre[presum]==2) st[top]=-st[top];
                presum--;
            }
            fh[top]=-1;now=next;
        }
        else
        {
            int End=word(now),all=0,end=End+1;
            while(code[end]!=')')
                pa[++all]=calc(end),end+=code[end]==',';
            PA=m2[code.substr(now,End-now)];retu=0;
            run(funstart[PA],1);
            st[++top]=retu;fh[top]=-1;isvar[top]=0;
            while(presum>0)
            {
                if(pre[presum]==1) st[top]=!st[top];
                if(pre[presum]==2) st[top]=-st[top];
                presum--;
            }
            now=end+1;
        }
        else
        if(code[now]=='(')
        {
            now++;
            st[++top]=calc(now);
            while(presum>0)
            {
                if(pre[presum]==1) st[top]=!st[top];
                if(pre[presum]==2) st[top]=-st[top];
                presum--;
            }
            isvar[top]=0;fh[top]=-1;now++;
        }
        else
        if(!top || fh[top]!=-1) 
        while(code[now]=='!' || code[now]=='-' || code[now]=='+')
        switch(code[now])
        {
            case'!':pre[++presum]=1;now++;break;
            case'-':pre[++presum]=2;now++;break;
            case'+':now++;break;
        }
        else
        {
            int newlevel=level(now);
            while(top>1 && dayu(fh[top-1],newlevel))
            switch(fh[--top])
            {
                case 0:var[st[top]]=st(top+1);st[top]=st(top+1);isvar[top]=0;break;
                case 1:st[top]=st(top)||st(top+1);isvar[top]=0;break;
                case 2:st[top]=st(top)&&st(top+1);isvar[top]=0;break;
                case 3:st[top]=st(top)^st(top+1);isvar[top]=0;break;
                case 4:st[top]=st(top)==st(top+1);isvar[top]=0;break;
                case 5:st[top]=st(top)!=st(top+1);isvar[top]=0;break;
                case 6:st[top]=st(top)<=st(top+1);isvar[top]=0;break;
                case 7:st[top]=st(top)>=st(top+1);isvar[top]=0;break;
                case 8:st[top]=st(top)1);isvar[top]=0;break;
                case 9:st[top]=st(top)>st(top+1);isvar[top]=0;break;
                case 10:st[top]=st(top)+st(top+1);isvar[top]=0;break;
                case 11:st[top]=st(top)-st(top+1);isvar[top]=0;break;
                case 12:st[top]=st(top)*st(top+1);isvar[top]=0;break;
                case 13:st[top]=st(top)/st(top+1);isvar[top]=0;break;
                case 14:st[top]=st(top)%st(top+1);isvar[top]=0;break;
            }
            fh[top]=newlevel;now++;
        }
    for(top--;top>0;top--)
    switch(fh[top])
    {
        case 0:var[st[top]]=st(top+1);st[top]=st(top+1);isvar[top]=0;break;
        case 1:st[top]=st(top)||st(top+1);isvar[top]=0;break;
        case 2:st[top]=st(top)&&st(top+1);isvar[top]=0;break;
        case 3:st[top]=st(top)^st(top+1);isvar[top]=0;break;
        case 4:st[top]=st(top)==st(top+1);isvar[top]=0;break;
        case 5:st[top]=st(top)!=st(top+1);isvar[top]=0;break;
        case 6:st[top]=st(top)<=st(top+1);isvar[top]=0;break;
        case 7:st[top]=st(top)>=st(top+1);isvar[top]=0;break;
        case 8:st[top]=st(top)1);isvar[top]=0;break;
        case 9:st[top]=st(top)>st(top+1);isvar[top]=0;break;
        case 10:st[top]=st(top)+st(top+1);isvar[top]=0;break;
        case 11:st[top]=st(top)-st(top+1);isvar[top]=0;break;
        case 12:st[top]=st(top)*st(top+1);isvar[top]=0;break;
        case 13:st[top]=st(top)/st(top+1);isvar[top]=0;break;
        case 14:st[top]=st(top)%st(top+1);isvar[top]=0;break;
    }
    if(code[now]==';')
        now++;
    return st(1);
}
int sentence(int now)//执行语句
{
    int run(int now,bool isfun);
    int next=word(now);
    if(code[now]=='{')
        next=run(now,0);
    else
    if(next-now==2 && code.substr(now,2)=="if")
    {
        newdep();
        next++;
        int end=express(next);
        if(calc(next))
        {
            next++;next=sentence(next);
            int end=word(next);
            if(end-next==4 && code.substr(next,4)=="else")
                next=block(end);
        }
        else
        {
            next++;
            next=block(next);
            int end=word(next);
            if(end-next==4 && code.substr(next,4)=="else")
                next=sentence(end+(code[end]==' '));
        }
        dep--;
    }
    else
    if(next-now==3 && code.substr(now,3)=="for")
    {
        newdep();
        int judge;
        judge=(code[next+1]==';')?next+2:sentence(next+1);int ju=judge;
        if(RETU) return next;
        int done=express(judge)+1;
        int blo=done,deep=1;
        while(deep)
        {
            if(code[blo]=='(') deep++;
            if(code[blo]==')') deep--;
            blo++;
        }
        while((code[ju]==';')?1:calc(ju))
        {
            ju=judge,sentence(blo);
            if(RETU) return next;
            sentence(done);
            if(RETU) return next;
        }
        next=block(blo);
        dep--;
    }
    else
    if(next-now==5 && code.substr(now,5)=="while")
    {
        newdep();
        int judge=next+1,ju=judge;
        if(RETU) return next;
        int blo=judge,deep=1;
        while(deep)
        {
            if(code[blo]=='(') deep++;
            if(code[blo]==')') deep--;
            blo++;
        }
        while(calc(ju))
        {
            sentence(blo),ju=judge;
            if(RETU) return next;
        }
        next=block(blo);
        dep--;
    }
    else
    if(next-now==3 && code.substr(now,3)=="cin")
    {
        int qian=next,hou=next+1;
        while(code[qian]!=';')
        {
            qian=word(hou+1);
            int first=m1[dep][code.substr(hou+1,qian-hou-1)];
            var[first+focus(qian,first)]=input[++read];
            hou=qian+1;
        }
        next=hou;
    }
    else
    if(next-now==4 && code.substr(now,4)=="cout")
    {
        int qian=next,hou=next+1;
        while(code[qian]!=';')
        {
            qian=hou+1;
            while((code[qian]!='<' || code[qian+1]!='<' )&&(code[qian]!=';')) qian++;
            if(qian-hou==5 && code.substr(hou+1,4)=="endl")
                puts("");
            else
                hou++,printf("%d",calc(hou));
            hou=qian+1;
        }
        next=hou;
    }
    else
    if(next-now==7 && code.substr(now,7)=="putchar")
        next++,printf("%c",calc(next)),next+=2;
    else
    if(next-now==3 && code.substr(now,3)=="int")
        next=newvar(next+1);
    else
    if(next-now==6 && code.substr(now,6)=="return")
        next++,retu=calc(next),RETU=1;
    else
    {
        calc(now);
        next=now;
    }
    return next;
}
int run(int now,bool isfun)//从某个{开始运行 
{
    newdep();now++;
    if(PA)
        for(int i=1;i<=pasum[PA];i++)
            m1[dep][par[PA][i]]=varsum[dep],var[varsum[dep]]=pa[i],varsum[dep]++;
    PA=0;
    for(int next=now;code[next]!='}'&& !RETU;now=next)
        next=sentence(now);
    dep--;if(isfun)RETU=0;now++;
    return now;
}
int main()
{
    init();
    for(int i=0,j;i//处理全局变量和函数 
    {
        i=word(i)+1;j=word(i);//过滤int
        if(code[j]=='(')    j=newfun(i,j);
        else    j=newvar(i);
    }
    run(funstart[m2["main"]],1);
    return 0;
 } 

牛逼的a+ b problem