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UVA 11517 Exact Change (01背包)

2015-11-13 01:00 274 查看
题意:

n<=100个不用面值的硬币组成大于等于w<=10000的最小值,并输出最少使用硬币个数

题意:

嘛,硬币不超过10000,最多多一个,所以背到20000就好了

代码:

//
//  Created by TaoSama on 2015-11-07
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m, w[105], dp[20005];

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
while(t--) {
scanf("%d%d", &m, &n);
for(int i = 1; i <= n; ++i) scanf("%d", w + i);
memset(dp, 0x3f, sizeof dp);
dp[0] = 0;
for(int i = 1; i <= n; ++i)
for(int j = 20000; j >= w[i]; --j)
dp[j] = min(dp[j], dp[j - w[i]] + 1);
int ans;
for(int i = m; i <= 20000; ++i)
if(dp[i] != INF) {ans = i; break;}
printf("%d %d\n", ans, dp[ans]);
}
return 0;
}
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标签:  01背包