POJ 2409

水题一道,不加优化也能0MS

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

int gcd(int a,int b){
	if(b==0) return a;
	return gcd(b,a%b);
}

int Power(int a,int b){
	int ans=1;
	while(b){
		if(b&1) ans=(ans*a);
		a=a*a;
		b>>=1;
	}
	return ans;
}

int main(){
	int c,s;
	while(scanf("%d%d",&c,&s),c||s){
		int ans=0;
		for(int i=1;i<=s;i++){
			ans=ans+Power(c,gcd(i,s));
		}
		if(s&1){
			ans=ans+s*Power(c,s/2+1);
		}
		else{
			ans=ans+(s/2)*(Power(c,s/2+1)+Power(c,s/2));
		}
		ans/=(2*s);
		printf("%d
",ans);
	}
	return 0;
}

  

原文地址:https://www.cnblogs.com/jie-dcai/p/4030632.html