| Time Limit: 2 second(s) | Memory Limit: 32 MB |
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 starts with an integer T (≤ 10000), denoting the number of test cases.
Each case contains an integer Q (1 ≤ Q ≤ 108) in a line.
For each case, print the case number and N. If no solution is found then print ‘impossible‘.
Sample Input |
Output for Sample Input |
|
3 1 2 5 |
Case 1: 5 Case 2: 10 Case 3: impossible |
题意：给你一个数Q，代表N！中末尾连续0的个数。让你求出最小的N。
求N！中尾连续0的个数：
LL change(LL x){
LL ans = 0;
while(x){
ans += x / 5;
x /= 5;
}
return ans;
}#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
#define INF 0x3f3f3f3f
#define LL long long
LL change(LL x){
LL ans = 0;
while(x){
ans += x / 5;
x /= 5;
}
return ans;
}
int main (){
int T, n;
int k = 1;
scanf("%d", &T);
while(T--){
scanf("%d", &n);
LL l = 0, r = 100000000 * 5 + 10;
LL mid, ans;
while(r > l){
mid = (l + r) / 2;
if(change(mid) >= n){
ans = mid;
r = mid;
}
else
l = mid + 1;
}
printf("Case %d: ", k++);
if(change(ans) == n)
printf("%lld\n", ans);
else
printf("impossible\n");
}
return 0;
}
版权声明：本文为博主原创文章，未经博主允许不得转载。
Light oj 1138 - Trailing Zeroes (III) 【二分查找 && N！中末尾连续0的个数】
原文地址：http://blog.csdn.net/hpuhjh/article/details/47816093
