【题解】洛谷 P2725 邮票 Stamps

目录

• [TOC]
  • 题目
  • 思路
  • $Code$

题目

P2725 邮票 Stamps

思路

\(\texttt{dp}\)。\(\texttt{dp[i]}\)表示拼出邮资\(i\)最少需要几张邮票。

状态转移方程：\(\texttt{dp[i]=min(dp[i],dp[i-value]+1)}\)

\(Code\)

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,m,a[51],f[2000001];
inline void read(int &T){
	int x=0;bool f=0;char c=getchar();
	while(c<'0'||c>'9'){if(c=='-')f=!f;c=getchar();}
	while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
	T=f?-x:x;
}

int main(){
	read(n),read(m);
	int maxx=0;
	memset(f,0x3f3f3f,sizeof(f));
	for(int i=1;i<=m;++i){
		read(a[i]);
		maxx=max(a[i],maxx);
	}
	int stop=maxx*n;
	f[0]=0;
	for(int j=1;j<=m;++j){
		for(int i=1;i<=stop;++i){
			if(a[j]>i) continue;
			f[i]=min(f[i],f[i-a[j]]+1);
		}
	}
	for(int i=1;i<=stop;++i){
		if(f[i]>n){
			printf("%d\n",i-1);
			return 0;
		}
	}
	printf("%d\n",stop);
	return 0;
}