Help Needed in problem candies from atcoder educational dp contest

Revision en2, by Normal_People, 2024-07-17 10:41:16

I solved this problem(with right logic)but i do not know how this implementation is working. I am trying to make a prefix sum in dp array but when you print it there is no prefix array in it.(IG it's because of mod function, but i do not know it's inner working).

#include <bits/stdc++.h>
using namespace std;
#define fr(i, n) for (int64_t i = 0; i < n; i++)
#define frn(i, a, n) for (int64_t i = a; i < n; i++)
#define testoutput FILE *file = fopen("test.out", "r"); if (file){fclose(file);freopen("test.out", "w", stdout); }
#define vet vector<int64_t>
#define vett vector<vector<int64_t> >
#define vst vector<string>
#define en cout << "\n";
#define all(v) v.begin(), v.end()
#define int int64_t
#define mp make_pair
#define F first
#define S second
const int mx = INT64_MAX;

// clang++ -std=c++20 check.cpp -o a
// clang++ -std=c++20 random.cpp -o r


const int MOD = (1e9 + 7);
inline int posmod(int k)
{
    return ((k % MOD) + MOD) % MOD;
}

void solve()
{
    int n,k;cin >> n >> k;
    vet a(n+1);frn(i,1,n+1)cin >> a[i];
    vett dp(n+1,vet(k+1,0));dp[0][0]=1;
    frn(i,1,n+1){
        fr(j,k+1){
            int start = j-a[i];
            if(start<=0)start=0;
            else start = dp[i-1][start-1];
            dp[i][j]=posmod(dp[i][j]+dp[i-1][j]-start);
            if(j>0)dp[i][j] = posmod(dp[i][j]+dp[i][j-1]);
        }
    }
    cout << dp[n][k];
}


int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
    testoutput
    int64_t t = 1,caseno = 1;
    // cin >> t; 
    while (t--)
    {
        // cout << "Case # " << caseno << ": "<<endl;
        solve();
        // cout << "\n";
        // caseno++;
    }
    // cerr << "Time elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
}
Tags dp, atcoder educational dp

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en2 English Normal_People 2024-07-17 10:41:16 171
en1 English Normal_People 2024-07-17 09:56:25 1744 Initial revision (published)