Question Name:ROY & FLOWER FARM

#include <bits/stdc++.h> 
#define int long long
#define fi first
#define se second
#define pb push_back
using namespace std; 
int mod=1e9+7;
int gcd(int a,int b){
 if(b==0)return a;
 return  gcd(b,a%b);
}
 
 
int32_t main(){
 ios_base::sync_with_stdio(false);
 int t;
 cin>>t;
 while(t--)
 {
  int n,p;
  cin>>n>>p;
  vector<pair<int,int> >v;
  v.pb({0,0});
  for(int i=1;i<=n;i++)
  {
      int x,y;
      cin>>x>>y;
      if(x-y>=0)
      {
          v.pb({x-y,y});
      }
  }
        n=v.size()-1;
  int dp[n+1][p+1];
  memset(dp,0,sizeof(dp));
 
  for(int i=1;i<=n;i++)
  {
   for(int j=1;j<=p;j++)
   {
    if(j>=v[i].se)
    {
     dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i].se]+v[i].fi);
    }
    else
        dp[i][j]=dp[i-1][j];
       //cout<<dp[i][j]<<" ";
   }//cout<<endl;
  }
  int spend=0,have=p;
  for(int i=0;i<=p;i++)
  {
      //cout<<dp[n][i]<<" ";
      if(dp[n][i]+p>have)
      {
          have=dp[n][i]+p;
          spend=i;
      }
  }//cout<<endl;
  cout<<spend<<" "<<have<<endl;
  //cout<<dp[n][p]<<endl;
  
 }
 
    return 0;
}

• Problem Description
  Roy is the owner of a flower plant farm and sells flower plants for a living. Now the problem with flower plants is that they wither in a night if not taken care of. So at the end of the day, to make flower plants wither-resistant Roy uses special fertilizers.
  
  There are N number of plants not sold on one particular day (say today), but Roy can sell them the next day (say tomorrow) if he fertilizes them tonight.
  
  Roy sells a flower plant for Rs. X. Cost of fertilizing a plant is Rs. Y. Currently Roy has P Rupees with him.
  
  Given the selling price X and fertilizing cost Y of N plants, your task is to find minimum amount A (A P) that he should spend in fertilizing plants such that the total money he has after selling plants on the next day is maximized. Assume that all the fertilized plants are sold on the next day.
  
  Input Format:
  
  First line contains integer T – number of test cases.
  Second line contains two space separated integers N and P.
  Following N lines each contain two space separated integers X and Y.
  
  Output Format:
  Print the two space separated integers A and B in a new line for each test case.
  
  Constraints:
  1<= T<= 10
  1 <= N <= 100
  1 <= X,Y <= 1000
  1<= P<= 10000
• Test Case 1
  Input (stdin)2
  2 50
  80 40
  60 20
  5 100
  100 40
  50 40
  40 50
  60 20
  100 50
  Expected Output20 90
  90 210
• Test Case 2
  Input (stdin)10
  4 56
  38 63
  54 11
  21 31
  55 65
  1 13
  89 67
  4 187
  24 11
  89 87
  97 22
  62 97
  3 100
  82 92
  85 71
  1 28
  4 233
  59 95
  4 66
  24 89
  24 32
  2 61
  15 52
  98 67
  2 97
  75 6
  19 94
  3 57
  3 41
  24 73
  36 17
  2 18
  12 98
  36 51
  2 69
  76 92
  38 95
  Expected Output11 99
  0 13
  120 277
  71 114
  0 233
  0 61
  6 166
  17 76
  0 18
  0 69

#include <bits/stdc++.h> 
#define int long long
#define fi first
#define se second
#define pb push_back
using namespace std; 
int mod=1e9+7;
int gcd(int a,int b){
 if(b==0)return a;
 return  gcd(b,a%b);
}
 
 
int32_t main(){
 ios_base::sync_with_stdio(false);
 int t;
 cin>>t;
 while(t--)
 {
  int n,p;
  cin>>n>>p;
  vector<pair<int,int> >v;
  v.pb({0,0});
  for(int i=1;i<=n;i++)
  {
      int x,y;
      cin>>x>>y;
      if(x-y>=0)
      {
          v.pb({x-y,y});
      }
  }
        n=v.size()-1;
  int dp[n+1][p+1];
  memset(dp,0,sizeof(dp));
 
  for(int i=1;i<=n;i++)
  {
   for(int j=1;j<=p;j++)
   {
    if(j>=v[i].se)
    {
     dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i].se]+v[i].fi);
    }
    else
        dp[i][j]=dp[i-1][j];
       //cout<<dp[i][j]<<" ";
   }//cout<<endl;
  }
  int spend=0,have=p;
  for(int i=0;i<=p;i++)
  {
      //cout<<dp[n][i]<<" ";
      if(dp[n][i]+p>have)
      {
          have=dp[n][i]+p;
          spend=i;
      }
  }//cout<<endl;
  cout<<spend<<" "<<have<<endl;
  //cout<<dp[n][p]<<endl;
  
 }
 
    return 0;
}

• Problem Description
  Roy is the owner of a flower plant farm and sells flower plants for a living. Now the problem with flower plants is that they wither in a night if not taken care of. So at the end of the day, to make flower plants wither-resistant Roy uses special fertilizers.
  
  There are N number of plants not sold on one particular day (say today), but Roy can sell them the next day (say tomorrow) if he fertilizes them tonight.
  
  Roy sells a flower plant for Rs. X. Cost of fertilizing a plant is Rs. Y. Currently Roy has P Rupees with him.
  
  Given the selling price X and fertilizing cost Y of N plants, your task is to find minimum amount A (A P) that he should spend in fertilizing plants such that the total money he has after selling plants on the next day is maximized. Assume that all the fertilized plants are sold on the next day.
  
  Input Format:
  
  First line contains integer T – number of test cases.
  Second line contains two space separated integers N and P.
  Following N lines each contain two space separated integers X and Y.
  
  Output Format:
  Print the two space separated integers A and B in a new line for each test case.
  
  Constraints:
  1<= T<= 10
  1 <= N <= 100
  1 <= X,Y <= 1000
  1<= P<= 10000
• Test Case 1
  Input (stdin)2
  2 50
  80 40
  60 20
  5 100
  100 40
  50 40
  40 50
  60 20
  100 50
  Expected Output20 90
  90 210
• Test Case 2
  Input (stdin)10
  4 56
  38 63
  54 11
  21 31
  55 65
  1 13
  89 67
  4 187
  24 11
  89 87
  97 22
  62 97
  3 100
  82 92
  85 71
  1 28
  4 233
  59 95
  4 66
  24 89
  24 32
  2 61
  15 52
  98 67
  2 97
  75 6
  19 94
  3 57
  3 41
  24 73
  36 17
  2 18
  12 98
  36 51
  2 69
  76 92
  38 95
  Expected Output11 99
  0 13
  120 277
  71 114
  0 233
  0 61
  6 166
  17 76
  0 18
  0 69