C++ :

#include <bits/stdc++.h>

using namespace std;

const int N = 5e4 + 1;
// 强度 >= P 的最小花费
int t, f[101][N], p[101], c[101]; // fij: 从前 i 个里选择，费用为 j 的最大强度

int main() {
    cin >> t;
    while (t--) {
        int n, P, Q;
        cin >> n >> P >> Q;
        for (int i = 1; i <= n; i++) {
            cin >> p[i] >> c[i];
        }

        int x = 1e9;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= Q; j++) {
                f[i][j] = f[i - 1][j];
                if (j >= c[i])
                    f[i][j] = max(f[i][j], f[i - 1][j - c[i]] + p[i]);
                if (f[i][j] >= P && j < x)
                    x = j;
            }
        }
        if (x > Q) cout << "-1\n";
        else cout << x << endl;

    }
}