Charlie's Change
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 3720 | Accepted: 1125 |
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：给出想要买的东西的价格p，有四枚硬币面值分别是1，5，10，25，然后给出每一种硬币的个数，求买这个东西最多需要多少枚硬币
这题很不错，num[i][j]用来记录i状态下第j枚硬币的个数，dp[i]用来记录硬币的个数；要求的就是在硬币个数最多的情况下每一种硬币的个数
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio> using namespace std;
const int MAX = + ;
int dp[MAX],num[MAX][],c[],p;
int w[]={,,,};
void ZeroOnePage(int cost, int mount, int kind) //01背包传递参数有这一种硬币的数量和种类
{
for(int i = p; i >= cost; i--)
{
if(dp[i - cost] > - && dp[i] < dp[i - cost] + mount) //求枚数最多的情况，01的数量需要参数传递，完全背包就是1
{
dp[i] = dp[i - cost] + mount;
for(int j = ; j < ; j++)
num[i][j] = num[i - cost][j]; //更改这一状态下每一种硬币的数量
num[i][kind] += mount;
}
}
}
void CompletePage(int cost, int kind)
{
for(int i = cost; i <= p; i++)
{
if(dp[i - cost] > - && dp[i] < dp[i - cost] + )
{
dp[i] = dp[i - cost] + ;
for(int j = ; j < ; j++)
num[i][j] = num[i - cost][j];
num[i][kind] += ;
}
}
}
void MultiplePage(int cost,int mount,int kind)
{
if(cost * mount >= p)
{
CompletePage(cost, kind);
return;
}
int k = ;
while(k < mount)
{
ZeroOnePage(k * cost, k, kind);
mount = mount - k;
k <<= ;
}
if(mount > )
ZeroOnePage(mount * cost, mount, kind);
return ;
}
int main()
{
while(scanf("%d", &p) != EOF)
{
int sum = p;
for(int i = ; i < ; i++)
{
scanf("%d", &c[i]);
sum += c[i];
}
if(sum == )
break;
memset(num,,sizeof(num));
memset(dp,-,sizeof(dp));
dp[] = ;
for(int i = ; i < ; i++)
if(c[i])
MultiplePage(w[i],c[i],i);
if(dp[p] > )
{
printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",num[p][],num[p][],num[p][],num[p][]);
}
else
{
printf("Charlie cannot buy coffee.\n");
}
}
return ;
}