题意：给你n种价值不同的邮票，最大的不超过10000元，一次最多贴k张，求1到多少都能被表示出来？n≤50，k≤200。
题解：dp[i]表示i元最少可以用几张邮票表示，那么对于价值a的邮票，可以推出dp[j]=min(dp[j],dp[j-a] 1)。j从a到k*10000顺序枚举，因为类似于完全背包。
代码语言：javascript复制/*
TASK:stamps
LANG:C
*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#define INF 0x3f3f3f3f
using namespace std;
int n,k;
int dp[2000005];
int main(){
freopen("stamps.in","r",stdin);
freopen("stamps.out","w",stdout);
memset(dp,INF,sizeof dp);
dp[0]=0;
scanf("%d%d",&k,&n);
int a;
for(int i=1;i<=n;i ){
scanf("%d",&a);
for(int j=a;j<=k*10000;j )
dp[j]=min(dp[j],dp[j-a] 1);
}
for(int i=1;;i )
if(dp[i]>k){
printf("%dn",i-1);
break;
}
/*
Test 1: TEST OK [0.011 secs, 11992 KB]
Test 2: TEST OK [0.011 secs, 11992 KB]
Test 3: TEST OK [0.011 secs, 11992 KB]
Test 4: TEST OK [0.011 secs, 11992 KB]
Test 5: TEST OK [0.011 secs, 11992 KB]
Test 6: TEST OK [0.011 secs, 11992 KB]
Test 7: TEST OK [0.000 secs, 11992 KB]
Test 8: TEST OK [0.011 secs, 11992 KB]
Test 9: TEST OK [0.011 secs, 11992 KB]
Test 10: TEST OK [0.076 secs, 11992 KB]
Test 11: TEST OK [0.216 secs, 11992 KB]
Test 12: TEST OK [0.076 secs, 11992 KB]
Test 13: TEST OK [0.119 secs, 11992 KB]
*/
return 0;
}
