UVa 357 Let Me Count The Ways

 Let Me Count The Ways 

After making a purchase at a large department store, Mel's change was 17 cents. He received 1 dime, 1 nickel, and 2 pennies. Later that day, he was shopping at a convenience store. Again his change was 17 cents. This time he received 2 nickels and 7 pennies. He began to wonder ' "How many stores can I shop in and receive 17 cents change in a different configuration of coins? After a suitable mental struggle, he decided the answer was 6. He then challenged you to consider the general problem.

Write a program which will determine the number of different combinations of US coins (penny: 1c, nickel: 5c, dime: 10c, quarter: 25c, half-dollar: 50c) which may be used to produce a given amount of money.

Input

The input will consist of a set of numbers between 0 and 30000 inclusive, one per line in the input file.

Output

The output will consist of the appropriate statement from the selection below on a single line in the output file for each input value. The number m is the number your program computes, n is the input value.

There are m ways to produce n cents change.

There is only 1 way to produce n cents change.

Sample input

17 
11
4

Sample output

There are 6 ways to produce 17 cents change. 
There are 4 ways to produce 11 cents change. 
There is only 1 way to produce 4 cents change.

动态规划，与UVa 147（http://www.cnblogs.com/lzj-0218/p/3554664.html）基本相同。

用数组money[i]保存不同的硬币面值

设dp[x]表示x这么多钱的表示方法，状态转移方程为dp[x]+=dp[x-money[i]]

此题中硬币一共有5种面值，所以要从小到大进行5次动态规划

注意输入为0时输出为1

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int money[5]={1,5,10,25,50};

int n;
long long dp[30050];

int main()
{
    memset(dp,0,sizeof(dp));
    dp[0]=1;

    for(int i=0;i<5;i++)
        for(int j=money[i];j<=30000;j++)
            dp[j]+=dp[j-money[i]];

    while(scanf("%d",&n)==1)
    {
        if(dp[n]==1)
            printf("There is only 1 way to produce %d cents change.
",n);
        else
            printf("There are %lld ways to produce %d cents change.
",dp[n],n);
    }

    return 0;
}