[动态规划]UVA357 - 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

17 114

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.

题意：

经过在百货公司的一场血拼之后，小梅发现她身上的零钱共有17分（cent，美金货币单位，其他货币及面值请参考下方红字部分），分别是1个dime，1个nickel，以及2个penny。隔天，小梅去便利商店买完东西后发现她身上的零钱恰好又是17分，这次是2个nickel及7个penny。小梅就在想，有几种硬币的组合可能凑成17分呢？经过仔细算算之后，发现共有6种。

你的问题就是：给一个金额，请你回答共有多少种硬币组合的方式。

思路：动态规划的水题目，又是换硬币这一类型的题目。

#include<iostream>#include<cstring>using namespace std;const int maxn=30010;int val[]={1,5,10,25,50};long long dp[maxn];int main()    {        int i,j;        memset(dp,0,sizeof(dp));        dp[0]=1;        for(i=0;i<5;i++)            {                for(j=val[i];j<=maxn;j++)                    {                        dp[j]=dp[j]+dp[j-val[i]];                    }            }        int cnt;        while(cin>>cnt)            {                if(dp[cnt]>1) cout<<"There are "<<dp[cnt]<<" ways to produce "<<cnt<<" cents change."<<endl;                else cout<<"There is only 1 way to produce "<<cnt<<" cents change."<<endl;            }        return 0;    }