Codeforces 577B modulo Sum (dp| | Set) __codeforces

Topic Link:

Http://codeforces.com/problemset/problem/577/B

The main effect of the topic:

Give the number of N and ask if some of them are divisible by M.

Ideas:

When N>m, according to the pigeon Nest principle, can judge must be yes. Because for these numbers, after the modulus of the range is (0,m-1), then when the first m+1, at least after the number of modulo and the previous one has a repeat, at this time there must be two add can be divisible by M.

When n<=m, it is possible to consider using DP,DP[I][J] to represent the first I number modulo to obtain J.

Code:

#include <stdio.h>
#include <string.h>
int dp[1005][1005],a[1000005];
int main ()
{
    int n,m,i,j,k;
    while (scanf ("%d%d", &n,&m)!=eof)
    {
        memset (DP));
        for (i=1;i<=n;i++)
        {
          scanf ("%d", &a[i]);
          a[i]=a[i]%m;
    }
    if (n>m)
    {
        printf ("yes\n");
        Continue;
    }
  Dp[1][a[1]]=1;
    for (i=2;i<=n;i++)
    {
      dp[i][a[i]]=1;
        for (j=0;j<m;j++)
        {
            if (Dp[i-1][j]) {
                dp[i][j]=1;
                dp[i][(J+a[i])%m]=1
            }}
    if (Dp[n][0]) printf ("yes\n");
    else printf ("no\n");
    }

You can also use set to add the resulting and all to the set to see if the smallest one is 0.

#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <
Set> using namespace std;
int a[1000005],sum[100005];
      int main () {int n,m,i,j,k,flag;
      set<int>se;
          while (scanf ("%d%d", &n,&m)!=eof) {flag=0;
          memset (sum,0,sizeof (sum));
          Se.clear ();
          for (i=1;i<=n;i++) {scanf ("%d", &a[i]);
      a[i]=a[i]%m;
        } if (n>m) {printf ("yes\n");
      Continue
      } Set<int>::iterator it;
      Se.insert (a[1]);
          for (i=2;i<=n;i++) {it=se.begin ();
          int x=*it;
              if (x==0) {flag=1;
          Break
          int l=0;
             For (It=se.begin (); It!=se.end (); ++it) {int x=*it;
          Sum[l++]= (X+a[i])%m;
          for (j=0;j<l;j++) Se.insert (Sum[j]); Se.insert (a[i]);
      } it=se.begin ();
      if (*it==0) flag=1;
      if (flag) printf ("yes\n");
      else printf ("no\n");
return 0;
 }