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 mways to produce ncents change.

There is only 1 way to produce ncents 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.

题意：用1，5，10，25，50这五种面额去拼出我们需要的面值

思路：背包。。。

AC代码：

#include<stdio.h>

long long DP[33333];
int cent[5] = {1, 5, 10, 25, 50};

int main() {
    int n;
    while(scanf("%d", &n) != EOF) {
        for(int i = 0; i < 33333; i++)
            DP[i] = 1;
        for(int i = 1; i < 5; i++) {
            for(int j = 0; j < 33333; j++)
                DP[j + cent[i]] += DP[j];
        }
        if(DP[n] == 1)
            printf("There is only 1 way to produce %d cents change.\n", n);
        else
            printf("There are %lld ways to produce %d cents change.\n", DP[n], n);
    }
}

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原文：http://blog.csdn.net/u012797220/article/details/20959639