dfs出邮票的各种面值，然后dp求解。
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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define rep(i,n) for(int i=0;i<n;++i)
#define clr(x,c) memset(x,c,sizeof(x))
#define Rep(i,l,r) for(int i=l;i<=r;++i)
using namespace std;
const int maxn=40;
const int inf=5000000;
vector<int> ans;
int cur[maxn];
int ANS=0;
int ok[4000000];
int n,k;
int t=0;
void jud() {
int a=0; ok[0]=0;
while(ok[a]<=n) {
ok[++a]=inf;
rep(i,k) {
if(a<cur[i]) break;
ok[a]=min(ok[a],ok[a-cur[i]]+1);
}
}
if(--a>ANS) {
ANS=a;
ans.clear();
rep(i,k) ans.push_back(cur[i]);
}
}
void dfs(int x) {
if(x==k) jud();
else Rep(i,cur[x-1]+1,cur[x-1]*n+1) {
cur[x]=i;
dfs(x+1);
}
}
int main()
{
freopen("test.in","r",stdin);
freopen("test.out","w",stdout);
scanf("%d%d",&n,&k);
dfs(cur[0]=1);
rep(i,ans.size()) printf("%d ",ans[i]);
printf("\nMAX=%d\n",ANS);
return 0;
}
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1047 邮票面值设计
1999年NOIP全国联赛提高组
[NOIP1999提高] CODEVS 1047 邮票面值设计(dfs+dp)
原文：http://www.cnblogs.com/JSZX11556/p/4376571.html
