Your job is to help the children and present a solution.
InputThe input contains several test cases.
The first line of each test case contains two integers c and n (1 ≤ c ≤ n ≤ 100000), the number of children and the number of neighbours, respectively. The next line contains n space separated integers a 1 , ... , a n (1 ≤ a i ≤ 100000 ), where a i represents the number of sweets the children get if they visit neighbour i.
The last test case is followed by two zeros.
OutputFor each test case output one line with the indices of the neighbours the children should select (here, index i corresponds to neighbour i who gives a total number of a i sweets). If there is no solution where each child gets at least one sweet, print "no sweets" instead. Note that if there are several solutions where each child gets at least one sweet, you may print any of them.
Sample Input
4 5 1 2 3 7 5 3 6 7 11 2 5 13 17 0 0
Sample Output
3 5 2 3 4
题解: 鸽巢原理
参考代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstdio> 3 using namespace std; 4 const int maxn=1e5+10; 5 int c,n,a[maxn],temp[maxn]; 6 int main() 7 { 8 while(~scanf("%d%d",&c,&n),c+n) 9 { 10 for(int i=1;i<=n;i++) scanf("%d",a+i),temp[i]=0; 11 int sum=0; 12 for(int i=1;i<=n;i++) 13 { 14 sum=(sum+a[i])%c; 15 if(sum==0) 16 { 17 for(int j=1;j<i;j++) printf("%d ",j); 18 printf("%d ",i); 19 break; 20 } 21 else if(temp[sum]) 22 { 23 for(int j=temp[sum]+1;j<i;j++) printf("%d ",j); 24 printf("%d ",i); 25 break; 26 } 27 else temp[sum]=i; 28 } 29 } 30 return 0; 31 }