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Description:
You task is to find minimal natural number N, so that N! contains exactly Q zeroes on the trail in decimal notation. As you know N! = 1*2*...*N. For example, 5! = 120, 120 contains one zero on the trail.
Input:
Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each case contains an integer Q (1 ≤ Q ≤ 108) in a line.
Output:
For each case, print the case number and N. If no solution is found then print ‘impossible‘.
Sample Input:
3
1
2
5
Sample Output:
Case 1: 5
Case 2: 10
Case 3: impossible
题意：找阶乘尾0的个数是q的数，输出最小的那个。
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int Solve(int n) ///计算n的阶乘的尾0个数 { int sum = 0; while (n) { sum += n/5; n /= 5; } return sum; } int main () { int T, q, k = 0, i, mid, j, x, y; scanf("%d", &T); while (T--) { scanf("%d", &q); k++; x = -1; i = 1; j = 1e9; while (i <= j) ///用二分搜索枚举1e9以内的数的阶乘尾数0的个数，保存在y中 { mid = (i+j)/2; y = Solve(mid); if (y >= q) ///如果找到的尾数0的个数比q大，继续查找左边区间 { j = mid-1; if (y == q) ///如果找到的尾数0的个数等于q，保存这个值终止循环 { x = mid; break; } } else i = mid+1; ///如果找到的尾数0的个数比q小，继续查找右边区间 } if (x % 5 != 0) x = x-x%5; ///因为题目要求找到最小的x 所以要对x向下取整 ///(eg: q==3 时 二分查找到的x可能是15，16，17,18,19) if (x == 0) x = -1; ///可能找不到某个数的阶乘有q个0 if (x == -1) printf("Case %d: impossible\n", k); else printf("Case %d: %d\n", k, x); } return 0; }
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原文地址：http://www.cnblogs.com/syhandll/p/4746196.html
