POJ

 POJ  1787  Charlie's Change

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Description

Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at coffee vending machines at motorests. Charlie hates change. That is basically the setup of your next task.

Your program will be given numbers and types of coins Charlie has and the coffee price. The coffee vending machines accept coins of values 1, 5, 10, and 25 cents. The program should output which coins Charlie has to use paying the coffee so that he uses as many coins as possible. Because Charlie really does not want any change back he wants to pay the price exactly.

Input

Each line of the input contains five integer numbers separated by a single space describing one situation to solve. The first integer on the line P, 1 <= P <= 10 000, is the coffee price in cents. Next four integers, C1, C2, C3, C4, 0 <= Ci <= 10 000, are the numbers of cents, nickels (5 cents), dimes (10 cents), and quarters (25 cents) in Charlie's valet. The last line of the input contains five zeros and no output should be generated for it.

Output

For each situation, your program should output one line containing the string "Throw in T1 cents, T2 nickels, T3 dimes, and T4 quarters.", where T1, T2, T3, T4 are the numbers of coins of appropriate values Charlie should use to pay the coffee while using as many coins as possible. In the case Charlie does not possess enough change to pay the price of the coffee exactly, your program should output "Charlie cannot buy coffee.".

Sample Input

12 5 3 1 2
16 0 0 0 1
0 0 0 0 0

Sample Output

Throw in 2 cents, 2 nickels, 0 dimes, and 0 quarters.
Charlie cannot buy coffee.

题意:给你一个物品的价值　n　然后给你　4种硬币的数量（１　５　１０　２５这四种硬币）　问最多用多少硬币能正好凑够n　

一眼看去就是多重背包还原路径这也是第一种解法

//#include<bits/stdc++.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define inf 0x3f3f3f3f
int vis[10001][90];
int dp[10001];
int q[4];
int main()
{   freopen("in.txt","r",stdin);
    ios::sync_with_stdio(false);
    int n,c[4];
    int v[4]={1,5,10,25};
    while(cin>>n>>c[0]>>c[1]>>c[2]>>c[3]&&(n+c[1]+c[2]+c[3]+c[0]))
    {
        int a[101];
        int b[101];
        int d[101];
        int cnt=0;
        for(int i=0;i<4;i++)
        {
            int j=1;
            while(j<=c[i])
            {
                    a[cnt]=j*v[i];
                    b[cnt]=j;
                    d[cnt]=i;
                    cnt++;
                    c[i]-=j;
                    j*=2;
            }
            if(c[i])
            {
                    a[cnt]=c[i]*v[i];
                    b[cnt]=c[i];
                    d[cnt]=i;
                    cnt++;
            }


        }

        dp[0]=1;
        for(int i=0;i<cnt;i++)
        {

            for(int j=n;j>=a[i];j--)
            {
                if(dp[j-a[i]]&&dp[j]<dp[j-a[i]]+b[i])
                {
                    dp[j]=dp[j-a[i]]+b[i];
                    vis[j][i]=1;
                }

            }

        }
    //    cout<<dp[n]<<endl;
        if(!dp[n])
        {
            cout<<"Charlie cannot buy coffee."<<endl;
        }
        else
        {
            int j=n;
            for(int i=cnt-1;i>=0;i--)
            {
                if(vis[j][i])
                {  // cout<<d[i]<<endl;
                    q[d[i]]+=b[i];
                    j-=a[i];
                }

            }
            cout<<"Throw in "<<q[0]<<" cents, "<<q[1]<<" nickels, "<<q[2]<<" dimes, and "<<q[3]<<" quarters."<<endl;
        }
        for(int i=0;i<=n;i++)
        {
            dp[i]=0;
            for(int j=0;j<cnt;j++)
            vis[i][j]=0;
        }
        q[0]=q[1]=q[2]=q[3]=0;



    }



    return 0;
}

 第二种完全背包　记录使用次数　大于时不更新

写的有点乱，就这吧，这种方法时间短（自己还是想不起来．．）

//#include<bits/stdc++.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define inf 0x3f3f3f3f
int dp[10001];
int q[105];
int fa[10001];
int usr[10001];
int main()
{   //freopen("in.txt","r",stdin);
    ios::sync_with_stdio(false);
    int n,c[6];
    int v[9]={0,1,5,10,25};
    while(cin>>n>>c[1]>>c[2]>>c[3]>>c[4]&&(n+c[1]+c[2]+c[3]+c[4]))
    {
      memset(fa,0,sizeof(fa));
      memset(dp,0,sizeof(dp));
      memset(q,0,sizeof(q));
     dp[0]=1;
        for(int i=1;i<=4;i++)
        {
          memset(usr,0,sizeof(usr));
            for(int j=v[i];j<=n;j++)
            {

                if(dp[j-v[i]]&&dp[j]<dp[j-v[i]]+1&&usr[j-v[i]]<c[i])
                {
                    dp[j]=dp[j-v[i]]+1;
                    usr[j]=usr[j-v[i]]+1;
                    fa[j]=j-v[i];
                }

            }

        }
        if(!dp[n])
        {
            cout<<"Charlie cannot buy coffee."<<endl;
        }
        else
        {
            while(n!=0)
            {
                q[n-fa[n]]++;
                n=fa[n];

            }
            cout<<"Throw in "<<q[1]<<" cents, "<<q[5]<<" nickels, "<<q[10]<<" dimes, and "<<q[25]<<" quarters."<<endl;
        }

        //q[0]=q[1]=q[2]=q[3]=0;



    }



    return 0;
}