2012/8/9 关于今天的比赛

A 大数运算

大数模版写得不熟，然后出了不少错误，准备自己写一个模版。

 

G 拓展欧几里德求逆元

这道题公式没推错，代入参数搞错了，然后一直在调，浪费不少时间。

 

I 求线段交点

很普通的计算题，WA了6次，最后才A掉，计算时没有考虑double比较大小的细节，手写的fabs没过。

 

今天就A了三道最水的题，比较让我失望，I题不应该错这么多次，细节决定成败，这句话放在哪都没错，因为这样WA掉罚时真的很不值。希望明天比赛顺利，然后进入校队。

A：

#include
#include<string.h>
#include
#include
using namespace std;

const int base = 10000;
const int width = 4;
const int N = 11000;
char buf[11000];
struct bint
{
    int ln, v[N];
    bint(int r = 0){
        for(ln = 0; r > 0; r /= base) v[ln ++] = r % base;
    }
    bint& operator = (const bint& r){
        memcpy(this, &r, (r.ln + 1) * sizeof(int));
        return * this;
    }
};

bint a, b, c;

bool operator < (const bint& a, const bint& b){
    int i;
    if(a.ln != b.ln) return a.ln < b.ln;
    for(i = a.ln - 1; i >= 0 && a.v[i] == b.v[i]; i --);
    return i < 0 ? 0 : a.v[i] < b.v[i];
}

bool operator <= (const bint& a, const bint& b){
    return !(b < a);
}

bint operator + (const bint& a, const bint& b)
{
    bint res;
    int i, cy = 0;
    for(i = 0; i < a.ln || cy > 0; i ++){
        if(i < a.ln) cy += a.v[i];
        if(i < b.ln) cy += b.v[i];
        res.v[i] = cy % base;
        cy /= base;
    }
    res.ln = i;
    return res;
}

bint operator - (const bint& a, const bint& b){
    bint res;
    int i, cy = 0;
    for(res.ln = a.ln, i = 0; i < res.ln; i ++){
        res.v[i] = a.v[i] - cy;
        if(i < b.ln) res.v[i] -= b.v[i];
        if(res.v[i] < 0) cy = 1, res.v[i] += base;
        else cy = 0;
    }
    while(res.ln > 0 && res.v[res.ln - 1] == 0) res.ln --;
    return res;
}

bint operator * (const bint& a, const bint& b)
{
    bint res;res.ln = 0;
    if(0 == b.ln) {res.v[0] = 0; return res;}
    int i, j, cy;
    for(i = 0; i < a.ln; i ++){
        for(j = cy = 0; j < b.ln || cy > 0; j ++, cy /= base)
        {
            if(j < b.ln) cy += a.v[i] * b.v[j];
            if(i + j < res.ln) cy += res.v[i + j];
            if(i + j >= res.ln) res.v[res.ln ++] = cy % base;
            else res.v[i + j] = cy % base;
        }
    }
    return res;
}

int digits(bint& a)
{
    if(a.ln == 0) return 0;
    int l = (a.ln - 1) * 4;
    for(int t = a.v[a.ln - 1]; t; ++ l, t /= 10);
    return l;
}

bool read(bint& b, char buf[])
{
    if(1 != scanf("%s", buf)) return 0;
    int w, u, ln = strlen(buf);
    memset(&b, 0, sizeof(bint));
    if('0' == buf[0] && 0 == buf[1]) return 1;
    for(w = 1, u = 0; ln; ) {
        u += (buf[-- ln] - '0') * w;
        if(w * 10 == base){
            b.v[b.ln ++] = u; u = 0; w = 1;
        }
        else w *= 10;
    }
    if(w != 1) b.v[b.ln ++] = u;
    return 1;
}

void write(const bint & b){
    int i;
    printf("%d", b.ln == 0 ? 0 : b.v[b.ln - 1]);
    for(i = b.ln - 2; i >= 0; i --)
        printf("%04d", b.v[i]);
    printf("\n");
}

int main()
{
    char op[5];
    while(true)
    {
        if(!read(a, buf)) break;
        scanf("%s", op);
        memset(buf, 0, sizeof buf);
        read(b, buf);
        memset(&c, 0, sizeof(bint));
        if('+' == op[0])
        {
            c = a + b;
        }
        else if('-' == op[0])
        {
            c = a - b;
        }
        else if('*' == op[0])
        {
            c = a * b;
        }
        write(c);
    }
    return 0;
}

G：

// n * x - m * y = 1; d = gcd(n, m);

#include
#include<string.h>
#include

typedef long long LL;
LL n, m, x, y;

LL extgcd( LL a, LL b, LL &x, LL &y)
{
    if( b == 0) { x = 1; y = 0; return a;}
    LL d = extgcd( b, a % b, x, y);
    LL t = x;
    x = y;
    y = t - a / b * y;
    return d;
}

int main()
{
    while(scanf("%lld%lld", &n, &m) == 2)
    {
        LL d = extgcd(n, m, x, y);
        LL t = 1;
        if( t % d != 0)
        {
            printf( "No Solution.\n");
        }
        else{
            x = t / d * x;
            x = ( x % (m / d) + (m / d)) % (m / d);
            printf( "%lld\n", x) ;
        }

    }
    return 0;
}

I：

#include
#include

const double eps = 1e-8;

double x1, yy1, x2, y2, x3, y3, x4, y4;
double k1, k2, b1, b2;
double x, y;

int main()
{
    while(scanf("%lf%lf%lf%lf", &x1, &yy1, &x2, &y2) != EOF)
    {
        scanf("%lf%lf%lf%lf", &x3, &y3, &x4, &y4);
        if(fabs(x1 - x2) < eps || fabs(x3 - x4) < eps)
        {
            if(fabs(x1 - x2) < eps && fabs(x3 - x4) < eps)
            {
                printf("No Solution!\n");
                continue;
            }
            if(fabs(x1 - x2) < eps)
            {
                k2 = (y3 - y4) / (x3 - x4);
                b2 = y3 - k2 * x3;
                y = x1 * k2 + b2;
                x = x1;
            }

            if(fabs(x3 - x4) < eps)
            {
                k1 = (yy1 - y2) / (x1 - x2);
                b1 = yy1 - k1 * x1;
                y = x3 * k1 + b1;
                x = x3;
            }
            printf("%.6f %.6f\n", x, y);
            continue;
        }
        else
        {
            k1 = (yy1 - y2) / (x1 - x2);
            b1 = yy1 - k1 * x1;
            k2 = (y3 - y4) / (x3 - x4);
            b2 = y3 - k2 * x3;
            if(fabs(k1 - k2) < eps)
            {
                printf("No Solution!\n");
                continue;
            }
            x = (b2 - b1) / (k1 - k2);
            y = k1 * x + b1;
            printf("%.6f %.6f\n", x, y);
        }
    }
    return 0;
}

 

 

 

转载于:https://www.cnblogs.com/Yu2012/archive/2012/08/09/2630613.html