标签：51nod space nbsp show algorithm pre tput nod turn
2个数N和K，表示N个人，数到K出列。(2 <= N, K <= 10^6)
最后剩下的人的编号-----------------------------------------------------------------------------------------------------------------------------------------------------
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int phi(int n,int k){
if(n<=0) return 1;
return (phi(n-1,k)+k)%n;
}
int main(){
int n,k;
cin>>n>>k;
printf("%d\n",phi(n,k)+1);
return 0;
}
标签：51nod space nbsp show algorithm pre tput nod turn
原文地址：http://www.cnblogs.com/redips-l/p/7994922.html
