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[],aa,now,f[];
 map<int,bool> vis;

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