It's just a knapsack problem. Please think before asking.

Marinush

I guess it is a 2d dp question where while memoizing we'll take no. Of stones taken, sum.....where we will reduce sum value by stating a condition which shows that is sum > N make it equal to N only.....coz we'll care about the value of sum till then only

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define pi pair<int, int>

int mx;

int dp[1001][1001];

int f(vector<vector<int>> &arr,int idx,int rem)
{
    if(rem <= 0) return 0;
    if(idx >= arr.size()) return INT_MAX-mx;
    // take
    if(dp[idx][rem] != -1) return dp[idx][rem];
    int ans1 = arr[idx][1] + f(arr,idx+1,rem-1-arr[idx][0]);
    int ans2 = f(arr,idx+1,rem);

    return dp[idx][rem] = min(ans1,ans2);
}

void solve()
{
    memset(dp,-1,sizeof(dp));
    int n;
    cin>>n;
    mx = 0;
    int total = 0;
    vector<vector<int>> arr(n,vector<int>(2));
    for(int i=0;i<n;i++)
    {
        cin>>arr[i][0]>>arr[i][1];
        mx = max(mx,arr[i][1]);
        total += arr[i][1];
    }
    cout<<total-f(arr,0,n)<<"\n";
}

signed main()
{
    int t;
    t = 1;
    while (t--)
    {
        solve();
    }
}