2020牛客暑期多校训练营(第九场)

A. Groundhog and 2-Power Representation(eval应用)

传送门

题意:2(2)表示2^2 = 4,计算给出的表达式

eval函数计算表达式值,值得注意的是表达式中需要用**来表示乘方。一行过

搞了37分钟,python还是不熟啊

print(eval(str(input()).replace('(', '**(')))

F. Groundhog Looking Dowdy(尺取)

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题意:

共 n 天,每天有一些数量的衣服,每件衣服有个邋遢值,选取 m 天,m 天里每天选取一件衣服穿,最小化这 m 件衣服邋遢值的最大值和最小值之间的差值

思路:

把所有衣服按邋遢值排序,对应出每件衣服所代表的天数,假如区间 [l, r] 中共有 m 个不同的天数,则差值为 s[r].a - s[l].a,尺取解决,快读快写可优化200ms

尺取也不熟啊

scanf + printf:

+ 快读写:

 

#include 
using namespace std;
typedef long long ll;
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int N = 1e6 + 5;

inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while(!isdigit(ch)) {
        if(ch == '-')
            f = -1;
        ch = getchar();
    }
    while(isdigit(ch)) {
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    return x * f;
}

inline void write(int x)
{
    if(x < 0) x = -x, putchar('-');
    if(x > 9) write(x / 10);
    putchar(x % 10 + '0');
}

struct node
{
    int a, id;
} s[2 * N];

bool cmp(node x, node y) {
    return x.a <= y.a;
}

int mp[N];

int main()
{
    for(int i = 0; i < N; ++i) mp[i] = 0;
    int n, m, k, tot = 0;
    n = read(), m = read();
    for(int i = 1; i <= n; ++i) {
        k = read();
        while(k--) {
            s[++tot].a = read();
            s[tot].id = i;
        }
    }
    sort(s + 1, s + tot + 1, cmp);
    int l = 1, r = 0, cnt = 0;
    int minn = inf;
    while(1) {
        while(r < tot && cnt < m) {
            r++;
            if(!mp[s[r].id]) cnt++;
            mp[s[r].id]++;
        }
        if(cnt < m) break;
        minn = min(minn, s[r].a - s[l].a);
        mp[s[l].id]--;
        if(!mp[s[l].id]) cnt--;
        l++;
    }
    write(minn), printf("\n");
    return 0;
}

I. The Crime-solving Plan of Groundhog(思维 + 大数)

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题意:

给出 n 个数字(0~9),用它们构造两个没有前导零的正整数,求这两个数的最小积

思路:跟着感觉走,最小积肯定是拿出一个最小的数字来和其他的乘了,题解给的证明:

2020牛客暑期多校训练营(第九场)_第1张图片

套的模板,先处理出每个数字的个数再加,排序的话T了 

#include 
using namespace std;

typedef long long ll;
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const int N = 1e5 + 5;

//大数
struct BigInteger
{
    static const int BASE = 100000000; //和WIDTH保持一致
    static const int WIDTH = 8;        //八位一存储,如修改记得修改输出中的%08d
    bool sign;                         //符号, 0表示负数
    size_t length;                     //位数
    vector num;                   //反序存
    //构造函数
    BigInteger(long long x = 0)
    {
        *this = x;
    }
    BigInteger(const string &x)
    {
        *this = x;
    }
    BigInteger(const BigInteger &x)
    {
        *this = x;
    }
    //剪掉前导0,并且求一下数的位数
    void cutLeadingZero()
    {
        while (num.back() == 0 && num.size() != 1)
        {
            num.pop_back();
        }
        int tmp = num.back();
        if (tmp == 0)
        {
            length = 1;
        }
        else
        {
            length = (num.size() - 1) * WIDTH;
            while (tmp > 0)
            {
                length++;
                tmp /= 10;
            }
        }
    }
    //赋值运算符
    BigInteger &operator=(long long x)
    {
        num.clear();
        if (x >= 0)
        {
            sign = true;
        }
        else
        {
            sign = false;
            x = -x;
        }
        do
        {
            num.push_back(x % BASE);
            x /= BASE;
        }
        while (x > 0);
        cutLeadingZero();
        return *this;
    }
    BigInteger &operator=(const string &str)
    {
        num.clear();
        sign = (str[0] != '-'); //设置符号
        int x, len = (str.size() - 1 - (!sign)) / WIDTH + 1;
        for (int i = 0; i < len; i++)
        {
            int End = str.size() - i * WIDTH;
            int start = max((int)(!sign), End - WIDTH); //防止越界
            sscanf(str.substr(start, End - start).c_str(), "%d", &x);
            num.push_back(x);
        }
        cutLeadingZero();
        return *this;
    }
    BigInteger &operator=(const BigInteger &tmp)
    {
        num = tmp.num;
        sign = tmp.sign;
        length = tmp.length;
        return *this;
    }
    //绝对值
    BigInteger abs() const
    {
        BigInteger ans(*this);
        ans.sign = true;
        return ans;
    }
    //正号
    const BigInteger &operator+() const
    {
        return *this;
    }
    //负号
    BigInteger operator-() const
    {
        BigInteger ans(*this);
        if (ans != 0)
            ans.sign = !ans.sign;
        return ans;
    }
    // + 运算符
    BigInteger operator+(const BigInteger &b) const
    {
        if (!b.sign)
        {
            return *this - (-b);
        }
        if (!sign)
        {
            return b - (-*this);
        }
        BigInteger ans;
        ans.num.clear();
        for (int i = 0, g = 0;; i++)
        {
            if (g == 0 && i >= num.size() && i >= b.num.size())
                break;
            int x = g;
            if (i < num.size())
                x += num[i];
            if (i < b.num.size())
                x += b.num[i];
            ans.num.push_back(x % BASE);
            g = x / BASE;
        }
        ans.cutLeadingZero();
        return ans;
    }
    // - 运算符
    BigInteger operator-(const BigInteger &b) const
    {
        if (!b.sign)
        {
            return *this + (-b);
        }
        if (!sign)
        {
            return -((-*this) + b);
        }
        if (*this < b)
        {
            return -(b - *this);
        }
        BigInteger ans;
        ans.num.clear();
        for (int i = 0, g = 0;; i++)
        {
            if (g == 0 && i >= num.size() && i >= b.num.size())
                break;
            int x = g;
            g = 0;
            if (i < num.size())
                x += num[i];
            if (i < b.num.size())
                x -= b.num[i];
            if (x < 0)
            {
                x += BASE;
                g = -1;
            }
            ans.num.push_back(x);
        }
        ans.cutLeadingZero();
        return ans;
    }
    // * 运算符
    BigInteger operator*(const BigInteger &b) const
    {
        int lena = num.size(), lenb = b.num.size();
        BigInteger ans;
        for (int i = 0; i < lena + lenb; i++)
            ans.num.push_back(0);
        for (int i = 0, g = 0; i < lena; i++)
        {
            g = 0;
            for (int j = 0; j < lenb; j++)
            {
                long long x = ans.num[i + j];
                x += (long long)num[i] * (long long)b.num[j];
                ans.num[i + j] = x % BASE;
                g = x / BASE;
                ans.num[i + j + 1] += g;
            }
        }
        ans.cutLeadingZero();
        ans.sign = (ans.length == 1 && ans.num[0] == 0) || (sign == b.sign);
        return ans;
    }
    //*10^n 大数除大数中用到
    BigInteger e(size_t n) const
    {
        int tmp = n % WIDTH;
        BigInteger ans;
        ans.length = n + 1;
        n /= WIDTH;
        while (ans.num.size() <= n)
            ans.num.push_back(0);
        ans.num[n] = 1;
        while (tmp--)
            ans.num[n] *= 10;
        return ans * (*this);
    }
    // /运算符 (大数除大数)
    BigInteger operator/(const BigInteger &b) const
    {
        BigInteger aa((*this).abs());
        BigInteger bb(b.abs());
        if (aa < bb)
            return 0;
        char *str = new char[aa.length + 1];
        memset(str, 0, sizeof(char) * (aa.length + 1));
        BigInteger tmp;
        int lena = aa.length, lenb = bb.length;
        for (int i = 0; i <= lena - lenb; i++)
        {
            tmp = bb.e(lena - lenb - i);
            while (aa >= tmp)
            {
                str[i]++;
                aa = aa - tmp;
            }
            str[i] += '0';
        }
        BigInteger ans(str);
        delete[] str;
        ans.sign = (ans == 0 || sign == b.sign);
        return ans;
    }
    // %运算符
    BigInteger operator%(const BigInteger &b) const
    {
        return *this - *this / b * b;
    }
    // ++ 运算符
    BigInteger &operator++()
    {
        *this = *this + 1;
        return *this;
    }
    // -- 运算符
    BigInteger &operator--()
    {
        *this = *this - 1;
        return *this;
    }
    // += 运算符
    BigInteger &operator+=(const BigInteger &b)
    {
        *this = *this + b;
        return *this;
    }
    // -= 运算符
    BigInteger &operator-=(const BigInteger &b)
    {
        *this = *this - b;
        return *this;
    }
    // *=运算符
    BigInteger &operator*=(const BigInteger &b)
    {
        *this = *this * b;
        return *this;
    }
    // /= 运算符
    BigInteger &operator/=(const BigInteger &b)
    {
        *this = *this / b;
        return *this;
    }
    // %=运算符
    BigInteger &operator%=(const BigInteger &b)
    {
        *this = *this % b;
        return *this;
    }
    // < 运算符
    bool operator<(const BigInteger &b) const
    {
        if (sign != b.sign) //正负,负正
        {
            return !sign;
        }
        else if (!sign && !b.sign) //负负
        {
            return -b < -*this;
        }
        //正正
        if (num.size() != b.num.size())
            return num.size() < b.num.size();
        for (int i = num.size() - 1; i >= 0; i--)
            if (num[i] != b.num[i])
                return num[i] < b.num[i];
        return false;
    }
    bool operator>(const BigInteger &b) const
    {
        return b < *this;    // >  运算符
    }
    bool operator<=(const BigInteger &b) const
    {
        return !(b < *this);    // <= 运算符
    }
    bool operator>=(const BigInteger &b) const
    {
        return !(*this < b);    // >= 运算符
    }
    bool operator!=(const BigInteger &b) const
    {
        return b < *this || *this < b;    // != 运算符
    }
    bool operator==(const BigInteger &b) const
    {
        return !(b < *this) && !(*this < b);    //==运算符
    }
    // 逻辑运算符
    bool operator||(const BigInteger &b) const
    {
        return *this != 0 || b != 0;    // || 运算符
    }
    bool operator&&(const BigInteger &b) const
    {
        return *this != 0 && b != 0;    // && 运算符
    }
    bool operator!()
    {
        return (bool)(*this == 0);    // ! 运算符
    }

    //重载<<使得可以直接输出大数
    friend ostream &operator<<(ostream &out, const BigInteger &x)
    {
        if (!x.sign)
            out << '-';
        out << x.num.back();
        for (int i = x.num.size() - 2; i >= 0; i--)
        {
            char buf[10];
            //如WIDTH和BASR有变化,此处要修改为%0(WIDTH)d
            sprintf(buf, "%08d", x.num[i]);
            for (int j = 0; j < strlen(buf); j++)
                out << buf[j];
        }
        return out;
    }
    //重载>>使得可以直接输入大数
    friend istream &operator>>(istream &in, BigInteger &x)
    {
        string str;
        in >> str;
        size_t len = str.size();
        int start = 0;
        if (str[0] == '-')
            start = 1;
        if (str[start] == '\0')
            return in;
        for (int i = start; i < len; i++)
        {
            if (str[i] < '0' || str[i] > '9')
                return in;
        }
        x.sign = !start;
        x = str.c_str();
        return in;
    }
};

inline int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    return x*f;
}

int a[N], cnt[10];

int main()
{
	int t;
	t = read();
	while(t--) {
		int n;
		n = read();
		for(int i = 0; i < 10; ++i) cnt[i] = 0;
		for(int i = 0;i < n;i ++) {
			a[i] = read();
			cnt[a[i]]++;
		}
		BigInteger mi = 0;
		for(int i = 1; i < 10; ++i) {
            if(cnt[i]) {
                mi = i;
                cnt[i]--;
                break;
            }
		}
		string s = "";
		for(int i = 1; i < 10; ++i) {
            if(cnt[i]) {
                s += i + '0';
                cnt[i]--;
                break;
            }
		}
		for(int i = 0; i < 10; ++i)
            for(int j = 1; j <= cnt[i]; ++j)
                s += i + '0';
		BigInteger sum = s;
		BigInteger ans = mi * sum;
		cout<

 明天再补zzzzz

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