[POJ1742]Coins

最新推荐文章于 2020-05-19 12:34:42 发布

dxyinme

最新推荐文章于 2020-05-19 12:34:42 发布

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分类专栏： --------------------动态规划背包型动态规划

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背包型动态规划

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本文探讨了一种典型的支付组合问题，即如何利用不同面额的硬币支付从1到m之间的任意金额。通过动态规划的方法，文章详细介绍了如何计算出在给定的硬币种类和数量下，能够恰好支付的目标金额数目。

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Time Limit: 3000MS
Memory Limit: 30000K

Description

People in Silverland use coins.They have coins of value A1,A2,A3…An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn’t know the exact price of the watch.
You are to write a program which reads n,m,A1,A2,A3…An and C1,C2,C3…Cn corresponding to the number of Tony’s coins of value A1,A2,A3…An then calculate how many prices(form 1 to m) Tony can pay use these coins.

Input

The input contains several test cases. The first line of each test case contains two integers $n(1<=n<=100),m(m<=100000)$ .The second line contains 2n integers, denoting $A1,A2,A3...An,C1,C2,C3...Cn (1<=Ai<=100000,1<=Ci<=1000)$ . The last test case is followed by two zeros.

Output
For each test case output the answer on a single line.

Sample Input

3 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0

Sample Output

8
4

题意:
有 $n$ 种纸币，第 $i$ 种面值是 $Ai$ ,拥有的数量是 $Ci$ ,问能配出 $1到m$ 的中的几种面值。
第一行输入两个个数 $n，m$
第二行输入2n个数，前n个是 $A1...An$ ，后n个是 $C1...Cn$

题解:
楼教主男人八题之一….
$f[j]$ 表示面值 $j$ 是否能取到。
$g[i][j]$ 表示用第 $i$ 种纸币取到j的面值至少需要多少张
则转移方程转化为

if(!f[j] && f[j-a[i]] && g[i][j-a[i]] < c[i]){
    f[j]=1;
    g[i][j]=g[i][j-a[i]]+1;
}

$g$ 数组可以压掉 i <script type="math/tex" id="MathJax-Element-16">i</script>的那一维，每次做完记得要清零

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#define LiangJiaJun main
using namespace std;
int n,m,a[104],c[104];
int f[100004],g[100004];
int w33ha(){
    memset(f,0,sizeof(f));
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)scanf("%d",&c[i]);
    f[0]=1;
    int ans=0;
    for(int i=1;i<=n;i++){
        memset(g,0,sizeof(g));
        for(int j=a[i];j<=m;j++){
            if(!f[j]&&f[j-a[i]]&&g[j-a[i]]<c[i]){
                f[j]=1;
                g[j]=g[j-a[i]]+1;
            }
        }
    }
    for(int i=1;i<=m;i++)ans+=f[i];
    printf("%d\n",ans);
    return 0;
}

int LiangJiaJun(){
    while(scanf("%d%d",&n,&m)!=EOF){
        if(n==0&&m==0)break;
        w33ha();
    }
    return 0;
}