# Topcoder SRM 563 Div1 spellcards

Source: Internet
Author: User

Test instructions

There \ (n\) Zhang Fuka in a queue, each spell card has two attributes, Rank \ (li\) and damage \ (di\).
You can do any operation, each one of the following:

Move the team's first spell card to the end of the team.

Use the team's first spell card to damage the enemy ( d_i\) and discard the l_i\ Zhang Fuka (including the spell card you used). If the queue is insufficient \ (l_i\) Zhang Fuka Then you cannot use it.

The maximum value of the sum of the resulting damage is calculated.

\ (1<=n<=50, 1<=li<=50, 1<=di<=10000\)

Sol

Conclusion: If the chosen card satisfies \ (\sum l_i <= n\), then there is always a plan to play these cards

A backpack, van.

That's a pretty obvious conclusion, but it's not really that good in the exam room.

A simple proof:

The problem in the sequence is more difficult to handle, we put them on the ring, so as to eliminate the effect of operation 1

Obviously, the selected card must exist adjacent to the two (I-j > L_i\), that is, one of the release skills will not kill the other (otherwise must \ (>n\))

We used this card, so the problem turned into the same sub-problem.

#include<bits/stdc++.h>using namespace std;inline int read() {    char c = getchar(); int x = 0, f = 1;    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();    return x * f;}int f[100];class SpellCards{public:    int maxDamage(vector<int> l, vector<int> d) {        int N = l.size();        for(int i = 0; i < N; i++)             for(int j = N; j >= l[i]; j--)                f[j] = max(f[j], f[j - l[i]] + d[i]);        return *max_element(f + 1, f + N + 1);    }};int main() {    int N = read();    vector<int> l, d;    for(int i = 1; i <= N; i++) l.push_back(read());    for(int i = 1; i <= N; i++) d.push_back(read());    cout << SpellCards().maxDamage(l, d);}/*71 2 2 3 1 4 2113 253 523 941 250 534 454*/

Topcoder SRM 563 Div1 spellcards

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