POJ1787Charlie's Change

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Charlie‘s Change

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 2978		Accepted: 844

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 1 216 0 0 0 10 0 0 0 0

Sample Output

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

Source

題意：給定一個總金額，然後分別是1,5,10,25每種硬幣的數量，問最多可以拿多少硬幣恰好支付

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>const int MX=10000;using namespace std;int main(){    int i,j,k,g[4],v,c[4]={1,5,10,25},dp[11000][5];//dp[.][0]表示最多的總硬幣，dp[.][1~4]分別表示每種硬幣的數量    while(1)    {        int s=0;        scanf("%d",&v);        s+=v;        for(i=0;i<4;i++)        {            scanf("%d",g+i);            s+=g[i]*c[i];        }        //cout<<s<<endl;        if(s==0)break;        s-=v;        if(s<v)        {            printf("Charlie cannot buy coffee.\n");        }else if(s==v){            printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",g[0],g[1],g[2],g[3]);        }else{            memset(dp,-MX,sizeof dp);            for(i=0;i<5;i++)            dp[0][i]=0;            for(i=0;i<4;i++)            {                k=1;                while(g[i]-k>=0){                   for(j=v;j>=c[i]*k;j--)                   {                       if(dp[j][0]<dp[j-c[i]*k][0]+k)                       {                           dp[j][0]=dp[j-c[i]*k][0]+k;                           for(int x=1;x<5;x++)                           dp[j][x]=dp[j-c[i]*k][x];                           dp[j][i+1]+=k;                       }                   }                   g[i]-=k;                   k<<=1;                }                if(g[i]>0)                {                    k=g[i];                   for(j=v;j>=c[i]*k;j--)                   {                       if(dp[j][0]<dp[j-c[i]*k][0]+k)                       {                           dp[j][0]=dp[j-c[i]*k][0]+k;                           for(int x=1;x<5;x++)                           dp[j][x]=dp[j-c[i]*k][x];                           dp[j][i+1]+=k;                       }                   }                }            }            for(i=0;i<5;i++)            {                if(dp[v][i]<0)                {                    printf("Charlie cannot buy coffee.\n");break;                }            }            if(i==5){            printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",dp[v][1],dp[v][2],dp[v][3],dp[v][4]);            }        }    }    return 0;}