poj1787 Charlie's Change

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.

这题可以用多重背包，用dp[m][5]分别记录m价钱刚好能买到的物品个数的最大值，买1元，5元，10元，25元物品的个数。

#include<stdio.h>
#include<string.h>
#define inf 88888888
int max(int a,int b){
	return a>b?a:b;
}
int w[10]={0,1,5,10,25},v[10]={1,1,1,1,1,1};
int dp[10006][6];
int main()
{
	int n,m,i,j,a,b,c,d,k,ans,sum;
	int num[19];
	while(scanf("%d%d%d%d%d",&m,&num[1],&num[2],&num[3],&num[4])!=EOF)
	{
		if(m+num[1]+num[2]+num[3]+num[4]==0)break;
		if(m>num[1]+num[2]*5+num[3]*10+num[4]*25){
			printf("Charlie cannot buy coffee.\n");continue;
		}
		if(num[1]>=m){
			printf("Throw in %d cents, 0 nickels, 0 dimes, and 0 quarters.\n",m);continue;
		}
		for(i=0;i<=m;i++){
			dp[i][0]=-inf;
		}
		dp[0][0]=dp[0][1]=dp[0][2]=dp[0][3]=dp[0][4]=0;

		for(i=1;i<=4;i++){
			ans=num[i]*w[i];
			if(ans>=m){
				for(j=w[i];j<=m;j++){
					if(dp[j-w[i]][0]>=0){
						if(dp[j-w[i]][0]+v[i]>dp[j][0]){
							dp[j][0]=dp[j-w[i]][0]+v[i];
							dp[j][1]=dp[j-w[i]][1];dp[j][2]=dp[j-w[i]][2];dp[j][3]=dp[j-w[i]][3];dp[j][4]=dp[j-w[i]][4];
							dp[j][i]++;

						}
					}
				}
			}
			else{
				k=1;sum=0;
				while(sum+k<num[i]){
					sum+=k;
					for(j=m;j>=k*w[i];j--){
						if(dp[j-k*w[i]][0]>=0){
							if(dp[j-k*w[i]][0]+k*v[i]>dp[j][0]){
								dp[j][0]=dp[j-k*w[i]][0]+k*v[i];
							    dp[j][1]=dp[j-k*w[i]][1];dp[j][2]=dp[j-k*w[i]][2];dp[j][3]=dp[j-k*w[i]][3];dp[j][4]=dp[j-k*w[i]][4];
							    dp[j][i]+=k;

							}
						}

					}
					k=k*2;
				}
				k=num[i]-sum;
				for(j=m;j>=k*w[i];j--){
						if(dp[j-k*w[i]][0]>=0){
							if(dp[j-k*w[i]][0]+k*v[i]>dp[j][0]){
								dp[j][0]=dp[j-k*w[i]][0]+k*v[i];
							    dp[j][1]=dp[j-k*w[i]][1];dp[j][2]=dp[j-k*w[i]][2];dp[j][3]=dp[j-k*w[i]][3];dp[j][4]=dp[j-k*w[i]][4];
							    dp[j][i]+=k;

							}
						}

				}
			}
		}
		if(dp[m][0]>0)
		printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",dp[m][1],dp[m][2],dp[m][3],dp[m][4]);
		else printf("Charlie cannot buy coffee.\n");
		//printf("%d\n",dp[m][0]);
	}
	return 0;
}