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.

제목: 또 동전 문제.출력에는 몇 가지 구성 방식이 있다.
사고방식dp+완전 가방http://blog.csdn.net/accelerator_/article/details/10414651이것과 별 차이가 없다.
코드:

#include <stdio.h>
#include <string.h>

const int coin[5] = {1, 5, 10, 25, 50};
int n, i, j;
long long d[30005];
int main() {
	d[0] = 1;
	for (i = 0; i < 5; i ++)
		for (j = coin[i]; j <= 30000; j ++)
			d[j] += d[j - coin[i]];
	while (~scanf("%d", &n)) {
		if (d[n] == 1)
			printf("There is only %lld way to produce %d cents change.
", d[n], n);
		else
			printf("There are %lld ways to produce %d cents change.
", d[n], n);
	}
	return 0;
}