洛谷 P2725 邮票 Stamps

题目传送门

解题思路:

f[i]表示凑总面值i所需的最少邮票张数,然后快乐的跑完全背包.

AC代码:

#include<iostream>
#include<cstdio>
#include<map>
#include<cstring>

using namespace std;

int k,n,a[51],aa,now,f[2000001];
map<int,bool> vis;

int main() {
    scanf("%d%d",&k,&n);
    for(int i = 1;i <= n; i++) {
        scanf("%d",&a[i]);
        aa = max(aa,a[i]);
    }
    memset(f,0x3f3f3f,sizeof(f));
    f[0] = 0;
    for(int i = 1;i <= n; i++) f[a[i]] = 1;
    for(int i = 1;i <= n; i++)
        for(int j = a[i];j <= aa * k; j++)
            f[j] = min(f[j],f[j-a[i]] + 1);
    for(int i = 1;i <= aa * k; i++)
        if(f[i] > k) {
            printf("%d",i - 1);
            return 0;
        }
    printf("%d",k * aa);
    return 0;
}