标签：style blog color os io ar for 数据 div
How many zeros are there in the end of s! if both s and s! are written in base k which is not necessarily to be 10? For general base, the digit order is 0-9,A-Z,a-z(increasingly), for example F4 in base 46 is actually 694 in base 10,and f4 in base 46 is 1890 in base 10.
There are multiple cases(less than 10000). Each case is a line containing two integers s and k(0 ≤ s < 2^63, 2 ≤ k ≤ 62).
For each case, output a single line containing exactly one integer in base 10 indicating the number of zeros in the end of s!.
101 2 12 7
3 1
题意：给你s和k表示k进制的s；现在求s的10进制的阶乘换成k进制后末尾有几个0；
思路：找k进制的质因子个数kv[i]，以及s的阶乘中质因子的个数pn，那么最后换成k进制后有多少个0，就是pn/kv[i]中最小的。
刚开始找出来pn的时候，测试数据都过了，可是WA了，，，，后来我把ans开很大很大，结果就过了。郁闷
#include<cstdio> #include<cstring> #include<cmath> #include<iostream> #define N 65 using namespace std; char str[N]; int k; long long s; int prm[20]= {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61},kv[20]; /* int judge(int x,int tp) { int y=x,a=0; while(y<=tp) { a+=tp/y; // printf("y:%d\ttemp:%d ans:%d\n",y,temp,ans); if(tp/y<x) break; y*=x; } return a; }*/ void solve(long long x) { int i; long long res,ans=0x7ffffffffffffff;//开小了，，居然WA long long pn; for(i=0; i<18; i++) { res=x; res=res/prm[i]; pn=res; while(res) { res=res/prm[i]; pn+=res; } if(kv[i]&&pn/kv[i]<ans) ans=pn/kv[i]; } printf("%lld\n",ans); } int main() { int i,j; while(scanf("%s%d",str,&k)!=EOF) { int len=strlen(str); s=0; for(i=0; i<len; i++) { int num; if(str[i]<=‘z‘&&str[i]>=‘a‘) num=str[i]-‘a‘+36; else if(str[i]<=‘Z‘&&str[i]>= ‘A‘) num=str[i]-‘A‘+10; else num=str[i]-‘0‘; s=num+s*k; } memset(kv,0,sizeof(kv)); int n=k,x; for(i=0; i<18; i++) { while(n%prm[i]==0) { kv[i]++; n/=prm[i]; } } solve(s); } return 0; }
记得以前也做过类似的题。就是求阶乘中含多少质因子的题目。
Factorial Problem in Base K（zoj3621）
标签：style blog color os io ar for 数据 div
原文地址：http://www.cnblogs.com/yuyixingkong/p/3945337.html
