poj 1787 Charlie's Change 题解（动态规划）

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Charlie's Change

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 2865		Accepted: 794

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：给你一个整数p，告诉你价值为1,5,10,25的硬币的数量，让输出价值为p并且硬币数量最多的方案。

咋一看，就是一个多重背包，再记录一下路径就行了，但是将物品加上logn的优化以后任然超时，其实这个题有O（4*p）的算法。

首先：dp[ i ][ j ]，表示前i种硬币已经选了价值为j所用的硬币最多有多少个，use[ i ][ j ],表示dp[ i ][ j ]状态下第i种硬币用了多少个，如果我们从小到大的选择硬币，那么因为dp[ i ][ j ]已经保证了硬币是最多的，因为第i中硬币比前面的大，所以第i种硬币一定是能达到条件的情况下最小的，即，use[ i ][ j ]为状态dp[ i ][ j ]下第i种硬币最少使用多少个。那么状态转移方程就是：dp[ i ][ j ]=max(dp[ i-1 ][ j ],dp[ i-1 ][ j -xi]+1),dp[ i ][ j ]=max(dp[ i ][ j ],dp[ i ][ j-xi]+1)且(use[ i ][ j-xi] +1<=ci]),因为use[ i ][ j ]一定是最小的，所以这样转移一定是最优的。。。代码如下：

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<vector>
#include<algorithm>
#include<queue>
#include<stack>
#define nn 110
#define inff 0x3fffffff
#define mod 1000000007
#define eps 1e-9
using namespace std;
typedef long long LL;
int p;
int c[10];
int dp[5][11000];
int use[5][11000];
int qian[10];
int num[10];
int main()
{
    int i,j;
    qian[1]=1,qian[2]=5,qian[3]=10,qian[4]=25;
    while(scanf("%d%d%d%d%d",&p,&c[1],&c[2],&c[3],&c[4])!=EOF&&p+c[1]+c[2]+c[3]+c[4])
    {
        memset(use,0,sizeof(use));
        for(i=0;i<=4;i++)
        {
            for(j=0;j<=p;j++)
                dp[i][j]=-inff;
        }
        dp[0][0]=0;
        for(i=1;i<=4;i++)
        {
            for(j=0;j<=p;j++)
            {
                dp[i][j]=dp[i-1][j];
                if(j-qian[i]>=0)
                {
                    if(dp[i][j]<dp[i-1][j-qian[i]]+1&&c[i]>=1)
                    {
                        dp[i][j]=dp[i-1][j-qian[i]]+1;
                        use[i][j]=1;
                    }
                    if(dp[i][j]<dp[i][j-qian[i]]+1&&use[i][j-qian[i]]+1<=c[i])
                    {
                        dp[i][j]=dp[i][j-qian[i]]+1;
                        use[i][j]=use[i][j-qian[i]]+1;
                    }
                }
            }
        }
        if(dp[4][p]<0)
        {
            puts("Charlie cannot buy coffee.");
            continue;
        }
        memset(num,0,sizeof(num));
        int ix=4;
        int fc=p;
        while(ix)
        {
            num[ix]=use[ix][fc];
            fc-=qian[ix]*num[ix];
            ix--;
        }
        printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",num[1],num[2],num[3],num[4]);
    }
    return 0;
}