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.

Regionals 1990 >> North America - Southeast USA

问题链接：UVA367 UVALive5251 Let Me Count The Ways
问题简述：（略）
问题分析：
    0/1背包模板题。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA367 UVALive5251 Let Me Count The Ways */

#include <bits/stdc++.h>

using namespace std;

int coins[] = {1, 5, 10, 25, 50};
const int K = sizeof(coins) / sizeof(int);
const int N = 30000 + 1;
long long dp[N];

int main()
{
    memset(dp, 0, sizeof(dp));
    dp[0] = 1;
    for(int i = 0; i < K; i++)
        for(int j = coins[i]; j < N; j++)
            dp[j] += dp[j - coins[i]];

    int n;
    while(~scanf("%d", &n))
        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);

    return 0;
}