UVA 357 Let Me Count The Ways Problem（动态规划 硬币）

 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.

此题和 Coins change 类似，几乎是同一道题，但是呢，这个题让人郁闷的是在处理格式上。求支付一定面额有多少种方法。

注意点一个是输出要long long，一个是如果结果只有一种方法，输出格式跟其他的不同，在一点就是一定要切记，不要在最后位输出空格，即使你直接拷贝过来时后面有一个空格

还有，用滚动数组提交的时间是19ms，用母函数模板提交的是：463ms，孰优孰劣，显而易见

#include <iostream>
#include <cstdio>

using namespace std;
# define maxn 30002

long long  num[maxn];
long long  cent[6]={0,1,5,10,25,50};

int main()
{
    int i,t;
    long long  n;
    for( i=1;i<=maxn;i++)
        num[i]=0;
    num[0]=1;
    
    for(t=1;t<=5;t++)
        for(i=1;i<=maxn;i++)
        {
            if(i>=cent[t]) num[i]+=num[i-cent[t]];//  状态转移方程
        }
        while(scanf("%lld",&n)!=EOF)
        {
            if(num[n]==1)
                printf("There is only 1 way to produce %d cents change.\n",n);
            else
                printf("There are %lld ways to produce %lld cents change.\n",num[n],n);
        }
        return 0;
}