uva 357 Let Me Count The Ways

Let Me Count The Ways

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit

Status

Description




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.

状态转移方程：dp[j] += dp[j - a[i]]; a[i]是哪几种币值。数据比较大，dp数组用long long 。先打表，然后直接输出就行。

#include<stdio.h>
#include<string.h>
long long int  dp[30005];
int a[5] ={1,5,10,25,50};
int main()
{
memset(dp,0,sizeof(dp));
dp[0] = 1;
for(int i = 0;i < 5;i++)
for(int j = a[i];j <= 30000;j++)
dp[j] += dp[j - a[i]];
int x;
while(~scanf("%d",&x))
{
if(x < 5)
printf("There is only %d way to produce %d cents change.\n",1,x);
else
printf("There are %lld ways to produce %d cents change.\n",dp[x],x);
}
return 0;
}

Input

Output

Sample Input

Sample Output

Hint