阶乘因式分解(二) http://acm.nyist.net/JudgeOnline/problem.php?pid=70
2013-08-19 21:38
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阶乘因式分解(二)
时间限制:3000 ms | 内存限制:65535 KB难度:3
描述
[b][b]给定两个数n,m,其中m是一个素数。[/b][/b]
[b][b]将n(0<=n<=2^31)的阶乘分解质因数,求其中有多少个m。[/b][/b]
注:^为求幂符号。
输入第一行是一个整数s(0<s<=100),表示测试数据的组数
随后的s行, 每行有两个整数n,m。 输出输出m的个数样例输入
3 100 5 16 2 1000000000 13
样例输出
24 15 83333329
来源[苗栋栋]原创上传者苗栋栋
#include<stdio.h> int main() { int s; scanf("%d",&s); while(s--) { int i,n,m,a=0,b=0,c=0; int count=0; scanf("%d %d",&n,&m); while(n!=0) { count+=n/m; n=n/m; } printf("%d\n",count); } return 0; }
解决这道题的原理很简单,就是看看n里面有几个m,m*m,m*m*m...,
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