질문 L:Crested Ibis vs Monster----------------------------------------------------사유(완전 배낭)

제목 설명 Ibis is fighting with a monster.The health of the monster is H. Ibis can cast N kinds of spells. Casting the i-th spell decreases the monster’s health by Ai, at the cost of Bi Magic Points. The same spell can be cast multiple times. There is no way other than spells to decrease the monster’s health. Ibis wins when the health of the monster becomes 0 or below. Find the minimum total Magic Points that have to be consumed before winning.
Constraints ·1≤H≤104 ·1≤N≤103 ·1≤Ai≤104 ·1≤Bi≤104 ·All values in input are integers. Input is given from Standard Input in the following format 입력:
H N A1 B1 : AN BN
출력 Print the minimum total Magic Points that have to be consumed before winning.샘플 입력 Copy [샘플 1] 9 3 8 4 2 1 [샘플 2] 100 6 1 2 3 9 4 27 5 816 243 [샘플 3] 9999 10 540 7550 691 9680 700 9790 510 7150 415 5818 551 7712 587 8227 619 8671 8228 176 2461 샘플 출력 Copy [샘플 2] 4 [샘플 2] 100 [샘플 3] 139815 힌트 샘플 1 해석 First the first the first the monase the health'8 at the cost of 3 Magic Points. The monster’s health is now 1. Then, cast the third spell to decrease the monster’s health by 2, at the cost of 1 Magic Point. The monster’s health is now −1. In this way, we can win at the total cost of 4 Magic Points. 예제 2 설명 It is optimal to cast the first spell 100 times.
해석: 세트 문제, 완전 가방이지만 지정된 부피(가설 x)를 초과할 수 있기 때문에 우리의 부피는 x2로 옮겨간다.마지막으로 우리는 [x,x2] 구간에서 최소치를 찾았다

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

const int maxn = 2e4 + 10;

int n, m;
int v[maxn], w[maxn];
int f[maxn];

int main()
{
    scanf("%d%d",&n,&m);
    memset(f,0x3f,sizeof f);
    f[0]=0;
    for(int i = 1; i <= m; i ++)    scanf("%d%d",&v[i],&w[i]);
    for(int i = 1; i <= m; i ++){
        for(int j = v[i]; j <= 2e4; j ++)
            f[j] = min(f[j], f[j - v[i]] + w[i]);
    }
    int minn=0x3f3f3f3f;
    for(int i=n;i<=2e4;i++) minn=min(minn,f[i]);
    cout<<minn<<endl;
    return 0;
}