AKU NEGARAKU

Input will consist of a series of lines, each line containing the number of companies (N) with $1 \leq N \leq 1500$ , and a skipping value (M) with $1 \leq M \leq 50$ . The values will be terminated by a line consisting of double zeros (0 0) as shown in sample input output.

C++14(g++5.4) 解法, 执行用时: 16ms, 内存消耗: 396K, 提交时间: 2020-10-07 12:41:30

#include<iostream>
using namespace std;
int f(int n,int m){
	if(n==1) return 0;
	return (f(n-1,m)+m)%n;
}
int main(){
	int n,m;
	while(cin>>n>>m,n&&m){
		cout<<f(n,m)+1<<endl;
	}
	return 0;
}

pypy3(pypy3.6.1) 解法, 执行用时: 99ms, 内存消耗: 21592K, 提交时间: 2020-10-02 12:51:33

while True :
    n, m = map(int, input().split())
    if n == 0 and m == 0 :
        break
    ans = 0
    for i in range(2, n + 1) :
        ans = (ans + m) % i
    print(ans + 1)

C(clang 3.9) 解法, 执行用时: 13ms, 内存消耗: 256K, 提交时间: 2020-10-05 14:30:00

#include<stdio.h>
int main(){
	int m,n;
	while(scanf("%d%d",&m,&n)&&(m&&n)) {
		int s=0;
		for(int i=1;i<=m;i++){
			s=(s+n)%i;
		} 
	printf("%d\n",s+1);
		
	}
	return 0;
}

Python3(3.5.2) 解法, 执行用时: 147ms, 内存消耗: 3320K, 提交时间: 2020-10-02 12:37:42

N,M=map(int,input().split())
while not (N==0 and M==0):
    P=0
    for i in range(2,N+1):
        P=(P+M)%i
    print(P+1)
    N,M=map(int,input().split())

C++11(clang++ 3.9) 解法, 执行用时: 14ms, 内存消耗: 380K, 提交时间: 2020-10-02 12:02:21

#include<stdio.h>
int main()
{int i,x,jo,m;
while(~scanf("%d %d",&x,&m)&& x && m)
{

jo=0;
for(i=2;i<=x;i++)
jo=(jo+m)%i;
printf("%d\n",jo+1);
}
}

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