Knowledge Summary of poj 3370 pigeon cage principles

I heard about the principle of the drawer in the middle school. Unfortunately, I have never had the opportunity to see it. Now I have a conclusion on the principle of the pigeon cage. But I do not know the principle of the pigeon cage. I can summarize the principle of the pigeon cage first: if n + 1 items or n + 1 items are to be placed in n drawers, at least one drawer contains two or more items. If you know this conclusion: a1, a2, a3... am is a positive integer sequence with at least integers k and r, 1 <= k <r <= m, making ak + a (k + 1) +... + a (r) is a multiple of m. The proof is relatively simple: Sk indicates the sum of the number of the first k. (1) If Sk % m = 0, the number of the first k is a multiple of m (2) if Sn and St mode m are the same, the sum of the numbers from t + 1 to n is m equal to 0. even if you don't know this conclusion, if DP is so powerful, you should be able to think of using the first n concepts and the thinking. Conclusion: inference 1: m only pigeons, n cages, at least one pigeon Cage contains no less than m-1) /n] + 1 pigeon. Inference 2: If n m-1) + 1 balls are put into n boxes, at least one box has m balls. Inference 3: If m1, m2 ,... mn is n positive integers, and (m1 + m2 +... + mn)/n> R-1, m1, m2 ,... there must be at least one number of mn codes no less than r: no explanation, more than 700 MS. When judge was at that time, I was afraid of TLE.

#include<cstdio>  #include<cstring>  using namespace std;  #define N 100002    int sum[N],pos[N];    int main()  {        int c,n,i,r,t,j;        while(scanf("%d%d",&c,&n),c+n)      {          memset(pos,-1,sizeof(pos));          bool flag=false;            scanf("%d",&sum[0]);          sum[0]%=c;          pos[sum[0]]=0;          if(sum[0]==0){printf("1\n");flag=1;}            for(i=1;i<n;i++)          {              scanf("%d",&sum[i]);              if(flag)continue;              sum[i]%=c;              sum[i]+=sum[i-1];              sum[i]%=c;              if(sum[i]==0)              {                  for(j=0;j<i;j++)                      printf("%d ",j+1);                  printf("%d\n",i+1);                  flag=1;                  continue;              }              if(pos[sum[i]]==-1)pos[sum[i]]=i;              else              {                  for(j=pos[sum[i]]+1;j<=i;j++)                      if(j!=i)printf("%d ",j+1);                      else printf("%d\n",i+1);                  flag=1;              }          }                }        return 0;  }