CF 319 Div 2 Modulo Sum number theory +DP

　　　　　　　　　　　　　　　　　　Modulo Sum

Topic Abstract: Give you an array of integers, can you take out some trees, so that their and can be divisible by M.

Analysis: See code comments.

1#include <cstdio>2#include <cstring>3 using namespacestd;4 Const intMS =1000005;5 intN, M;6 intCnt[ms], R[ms], tem[ms];7 intMain () {8scanf"%d%d", &n, &m);9memset (CNT,0,sizeof(CNT));Ten      for(inti =0; I < n; i++) { Onescanf"%d", &r[i]); AR[i]%=m; -cnt[r[i]]++;//count the number of different residues -     } theMemset (R,0,sizeof(r));//initializing the remainder I cannot constitute -      for(inti =0; I < m; i++) { -         if(Cnt[i] >0) {//processing remainder I -             intx =1; +              while(Cnt[i] >0) { -                 inty = x < Cnt[i]?X:cnt[i]; +Cnt[i]-=y;  AX *=2;//multiplication algorithm processing.  at                 intz = (i * y)% m;//The y remainder i forms the remainder Z. -                  for(intj =0; J < M; J + +) { -                     if(R[j]) -tem[(j + z)% m] =1;//The remainder is formed on the basis of the remainder J (j + z)%m -                 } -                  for(intj =0; J < M; J + +) { inR[J] + = tem[j];//the remainder that the update can make -TEM[J] =0; to                 } +R[Z] =1;//The remainder Z can constitute -             } the             if(r[0] >0) *                  Break; $         }Panax Notoginseng     } -     if(r[0] >0) theprintf"yes\n"); +     Else Aprintf"no\n"); the     return 0; +}

CF 319 Div 2 Modulo Sum number theory +DP