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uva357 Let Me Count The Ways

注意PE……

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
int coin[5]={1,5,10,25,50};
long long dp[30010];
int main()
{
	int i,j;
	dp[0]=1;
	for(i=0;i<5;i++)
		for(j=coin[i];j<=30000;j++)
			dp[j]+=dp[j-coin[i]];
	while(cin>>i)
		if(dp[i]==1)
			printf("There is only 1 way to produce %d cents change.\n",i);
		else
			printf("There are %lld ways to produce %d cents change.\n",dp[i],i);
	return 0;
}

 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 numberm 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.


uva357 Let Me Count The Ways