15486 퇴사2
코드
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#include <iostream>
using namespace std;
int N, T[1500000], P[1500000], DP[1500000];
int dp(int deep);
int max(int a, int b) {
return a > b ? a : b;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> N;
for (int i = 0; i < N; i++) {
cin >> T[i] >> P[i];
}
cout << dp(0);
}
int dp(int deep) {
if (deep >= N)
return 0;
if (DP[deep] != 0)
return DP[deep];
int result = 0;
if (deep + T[deep] <= N)
result = dp(deep + T[deep]) + P[deep];
result = max(result, dp(deep + 1));
return DP[deep] = result;
}