poj 3370

和2356思路几乎相同，注意本题能使用的条件是因为c <= n

//============================================================================
// Name        : 3370.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

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

int n, num;
long long sum[100010];
int a[100010], flag[100010], from[100010];

int main() {
	while(scanf("%d%d", &n, &num)&&(n||num)){
		memset(flag, 0, sizeof(flag));
		for(int i = 1;i <= num;i++){
			scanf("%d\n", &a[i]);
			sum[i] = sum[i-1]+a[i];
		}
		for(int i = 1;i <= num;i++){
			if(sum[i]%n == 0){
				for(int j = 1;j < i;j++){
					printf("%d ", j);
				}
				printf("%d\n", i);
				break;
			}
			else{
				if(flag[sum[i]%n] == 0){
					flag[sum[i]%n] = 1;
					from[sum[i]%n] = i;
				}
				else{
					for(int j = from[sum[i]%n]+1;j < i;j++){
						printf("%d ", j);
					}
					printf("%d\n", i);
					break;
				}
			}
		}
	}
	return 0;
}