lightoj 1138 - Trailing Zeroes (III)【二分】

题目链接：http://lightoj.com/volume_showproblem.php?problem=1138

题意：问 N！ 末尾 0 的个数为 Q 个的数是什么?

解法：二分枚举Ｎ，因为０是由５×2 出现的，2的个数比5多故计算5的个数即可。

代码：

#include <stdio.h>#include <ctime>#include <math.h>#include <limits.h>#include <complex>#include <string>#include <functional>#include <iterator>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <set>#include <map>#include <list>#include <bitset>#include <sstream>#include <iomanip>#include <fstream>#include <iostream>#include <ctime>#include <cmath>#include <cstring>#include <cstdio>#include <time.h>#include <ctype.h>#include <string.h>#include <assert.h>using namespace std;long long count_num(long long n){    long long num = 0;    while (n)    {        num += n / 5;        n /= 5;    }    return num;}int main(){    int t;    scanf("%d",&t);    int cases = 1;    while (t--)    {        int n;        scanf("%d",&n);        printf("Case %d: ",cases++);        long long left = 1;        long long right = 1000000000000;        long long mid;        while (left <= right)        {            mid = (right + left) / 2;            long long tmp = count_num(mid);            if (tmp >= n)                right = mid - 1;            else                left = mid + 1;        }        if (count_num(left) != n)            puts("impossible");        else            printf("%lld\n",left);    }    return 0;}