完全背包问题UVA147

最新推荐文章于 2020-01-23 23:02:26 发布

Vict0riaWang

最新推荐文章于 2020-01-23 23:02:26 发布

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本文介绍了一个用于计算特定金额下不同货币组合数量的程序。该程序考虑了新西兰货币单位，包括纸币和硬币的各种面额，并通过动态规划算法来找出所有可能的组合方式。

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要考试了，实验班一直是我的愿望，虽然现在知道自己估计是上不了

可还是想试试

毕竟努力了那么久

<table bgcolor="#0060f0"><tbody><tr><td><strong><span style="font-size:24px; color:#c0ffff"><a target=_blank name="SECTION0001000000000000000000" target="_blank">Dollars</a></span> </strong></td></tr></tbody></table><p>New Zealand currency consists of $100, $50, $20, $10, and $5 notes and $2, $1, 50c, 20c, 10c and 5c coins. Write a program that will determine, for any given amount, in how many ways that amount may be made up. Changing the order of listing does not increase the count. Thus 20c may be made up in 4 ways: 1 <img width="9" height="18" align="MIDDLE" alt="tex2html_wrap_inline25" src="http://uva.onlinejudge.org/external/1/147img1.gif" /> 20c, 2 <img width="9" height="18" align="MIDDLE" alt="tex2html_wrap_inline25" src="http://uva.onlinejudge.org/external/1/147img1.gif" /> 10c, 10c+2 <img width="9" height="18" align="MIDDLE" alt="tex2html_wrap_inline25" src="http://uva.onlinejudge.org/external/1/147img1.gif" /> 5c, and 4 <img width="9" height="18" align="MIDDLE" alt="tex2html_wrap_inline25" src="http://uva.onlinejudge.org/external/1/147img1.gif" /> 5c.</p><p></p><h2><a target=_blank name="t0"></a><span style="color:#0070e8"><a target=_blank name="SECTION0001001000000000000000" target="_blank">Input</a></span></h2><p></p><p></p><p>Input will consist of a series of real numbers no greater than $300.00 each on a separate line. Each amount will be valid, that is will be a multiple of 5c. The file will be terminated by a line containing zero (0.00).</p><p></p><h2><a target=_blank name="t1"></a><span style="color:#0070e8"><a target=_blank name="SECTION0001002000000000000000" target="_blank">Output</a></span></h2><p></p><p></p><p>Output will consist of a line for each of the amounts in the input, each line consisting of the amount of money (with two decimal places and right justified in a field of width 6), followed by the number of ways in which that amount may be made up, right justified in a field of width 17.</p><p></p><h2><a target=_blank name="t2"></a><span style="color:#0070e8"><a target=_blank name="SECTION0001003000000000000000" target="_blank">Sample input</a></span></h2><p></p><p></p><p></p><pre>0.20
2.00
0.00

Sample output

  0.20                4
  2.00              293

#include<iostream>
#include<memory.h>
#include<stdio.h>
#include<string.h>
#define ll long long int
ll b[101],dp[11][10000];
int n;
int mon[11]={5,10,20,50,100,200,500,1000,2000,5000,10000};
int i;
float a[101];

using namespace std;
/*
ll DP(int  x,ll y)
{
        int m,k,h;
        if(dp[x][y]!=-1)
                return dp[x][y];
        dp[x][y]=0;
                for(m=0;y-m*mon[x]>=0;m++)
        {
                dp[x][y]+=DP(x-1,y-m*mon[x]);


        }
        return dp[x][y];
}
*/
ll DP(int  x,ll y)
{
        int m,k,h;
        if(dp[x][y]!=-1)
                return dp[x][y];
        dp[x][y]=0;
                for(m=0;y-m*mon[x]>=0;m++)
        {
                dp[x][y]+=DP(x-1,y-m*mon[x]);


        }
        return dp[x][y];
}
int main()
{
        dp[0][0]=0;
        int k=0;
        cin>>a[k];
        while(1)
        {

                memset(dp,-1,sizeof(dp));
                n=(int )(a[k]*100+0.5);
                //cout<<n;
                if(n==0)
                        break;
                        for(i=0;i<=n;i++)
                        {
                                dp[0][i]=1;
                        }
                       // cout<<a[k];
                b[k]=DP(10,n);
                k++;
        cin>>a[k];

        }
        for(i=0;i<k;i++)
        {
                printf("%.2f %d",a[i],b[i]);
        }
}

/*
#include<stdio.h>
#define N 30005
#include<string.h>
#define ll long long
int d[11]= {5,10,20,50,100,200,500,1000,2000,5000,10000};
ll dp[15][N];//dp[i][j]表示用前i+1种硬币，组成j分的种类数
ll DP(ll i,ll j)
{
    //printf("%d %d %d\n",i,j,dp[i][j]);
    if(dp[i][j]!=-1)
        return dp[i][j];
    dp[i][j]=0;
    for(int k=0;j-k*d[i]>=0;k++)
    {
        dp[i][j]+=DP(i-1,j-k*d[i]);
    }
    return dp[i][j];
}
int main()
{
    double nn;
    memset(dp,-1,sizeof(dp));//赋值一次即可，否则可能会超时
    while(scanf("%lf",&nn)!=EOF)
    {
        if(nn==0.00)
            break;
        int n=(int)(nn*100+0.5);//注意精度
        for(ll i=0; i<=n; i++)
            dp[0][i]=1;
            printf("%6.2f%17lld\n",nn,DP(10,n));
    }
    return 0;
}
/*
0.20
2.00
0.00
  0.20                4
  2.00              293
*/