lightoj-1138-Trailing Zeroes (III) -二分查找

题目传送门：https://vjudge.net/problem/LightOJ-1138

题意：给你一个数字，这个数字代表N！后面有几个0。给出这个数字，计算N的值。

解题思路：
任何质因数都可以写成素数相乘的形式。所以计算一个数的阶乘后面几个0，只需计算这个数包含多少5即可。

ac代码;

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <cctype>#include <string>#include <vector>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>#define inf 0x3f3f3f3f#define N 500000000typedef long long ll;using namespace std; int ok(int x){    int sum = 0;    while(x)    {        sum = sum + x/5;        x /= 5;    }    return sum;} int main(){    int T;    scanf("%d", &T);    for(int i = 1; i <= T; i++)    {        int n;        scanf("%d", &n);         int l = 0, r = N, mid;        while(l <= r)        {            mid = (l + r) /2;            int b = ok(mid);            if(b >= n)            {                r = mid-1;            }            else            {                l = mid+1;            }        }        if(ok(l) == n)            printf("Case %d: %d\n", i, l);        else            printf("Case %d: no\n", i);    }    return 0;}