标签：pos scan closed int cte ble onclick const can
题目链接：http://lightoj.com/volume_showproblem.php?problem=1138
题意：给你一个数n，然后找个一个最小的数x，使得x!的末尾有n个0；如果没有输出impossible
可以用二分求结果，重点是求一个数的阶乘中末尾含有0的个数，一定和因子5和2的个数有关，因子为2的明显比5多，所以我们只需要求一个数的阶乘的因子中一共有多少个5即可;
LL Find(LL x) { LL ans = 0; while(x) { ans += x/5; x /= 5; } return ans; }
例如125! 125/5 = 25；所以125!中含5的项为25*5+24*5+23*5+...+3*5+2*5+1*5 = 25!*5
所以125!末尾有25+5+1 = 31 个0;
#include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> typedef long long LL; #define N 1000001 using namespace std; const double eps = 1e-6; LL Find(LL x) { LL ans = 0; while(x) { ans += x/5; x /= 5; } return ans; } int main() { int T, t = 1, n; scanf("%d", &T); while(T --) { scanf("%d", &n); LL ans = 0, L = 2, R = 1000000000; while(L <= R) { LL Mid = (L+R)/2; LL ret = Find(Mid); if(ret == n) ans = Mid; if(ret >= n) R = Mid - 1; else L = Mid + 1; } if(ans == 0) printf("Case %d: impossible\n", t++); else printf("Case %d: %lld\n", t++, ans); } return 0; }
LightOj 1138 - Trailing Zeroes (III) 阶乘末尾0的个数 & 二分
标签：pos scan closed int cte ble onclick const can
原文地址：http://www.cnblogs.com/zhengguiping--9876/p/6007901.html
