Trailing Zeroes (III)【N!後0的個數&&二分(好題)】
阿新 • • 發佈:2019-02-07
1138 - Trailing Zeroes (III)
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
Input starts with an integer T (≤ 10000)
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 |
Output for Sample Input |
3 1 2 5 |
Case 1: 5 Case 2: 10 Case 3: impossible |
隨著N的增大 0的個數非遞減 二分列舉判斷即可
程式碼中統計N!中5的個數依然可以用於統計>1的任何整數;
AC程式碼:
#include<cstdio> typedef long long LL; const LL INF=1e18; LL Sum(LL N) {//N!後0的個數即統計N!能除以多少個5 直接統計1-N中5的倍數 LL ret=0; //考慮到1-N中有可能含有5的次方 所以縮小5倍之後繼續統計 while(N) { //即依次統計能被5^1整除的個數 5^2整除的個數 5^3 ..... N/=5; ret+=N; } return ret; } int main() { LL T,Kase=0; scanf("%lld",&T); while(T--) { LL Q; scanf("%lld",&Q); LL R=INF,L=1,ans=INF; while(L<=R) { LL mid=L+R>>1; LL s=Sum(mid); if(s==Q) { ans=mid;R=mid-1; } else if(s>Q) R=mid-1; else L=mid+1; } printf("Case %d: ",++Kase); if(ans>=INF) puts("impossible"); else printf("%lld\n",ans); } return 0; }