Cf-366C-Dima and Salad

Redefine the problem as c = a-B * k; find the maximum value of a when a series of c values are 0. Because c can be positive or negative. So I put the c positive 01 backpack once. Compress the 0-negative c package once. When the two are equal, add the aplus to the maximum value.

#include<stdio.h>  #include<string.h>  #include<algorithm>  #include<iostream>  using namespace std;  #define INF 99999999  int n,k,i,j;  int num[101];  int dp1[100001];  int dp2[100001];  int as[1001];  int bs[1001];  int main()  {      int a,b;      while(~scanf("%d%d",&n,&k))      {          for(i=0;i<=100000;i++)dp1[i]=dp2[i]=-INF;          dp1[0]=dp2[0]=0;          for(i=1;i<=n;i++)scanf("%d",&as[i]);          for(j=1;j<=n;j++)scanf("%d",&bs[j]);          for(i=1;i<=n;i++)          {              a=as[i];              b=bs[i];              num[i]=a-b*k;              if(num[i]<0)              {                  num[i]=-num[i];                  for(j=100000;j>=num[i];j--)                  {                      dp2[j]=max(dp2[j],dp2[j-num[i]]+a);                  }              }              else              {                  for(j=100000;j>=num[i];j--)                  {                      dp1[j]=max(dp1[j],dp1[j-num[i]]+a);                  }              }          }          int maxn;          maxn=-1;          for(i=10000;i>=0;i--)          {              maxn=max(maxn,dp1[i]+dp2[i]);          }          if(maxn==0)maxn=-1;          cout<<maxn<<endl;      }      return 0;  }