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.
以加入不同的面值來劃分決策階段,則種類數目等於之前的加上,加入此面之後增加的數目;
加入面值b[I]; 種類數a[I][J]=a[I][J-1]+a[I-b[I]][J]; 0 1 2 3 4 5 6 7 8 9 10 15 20 25 0 1 1 1 1 1 1 1 1 1 1 1 5 1 1 1 1 2 2 2 2 2 3 4 5 6 10 25 50
#include<stdio.h>
#include<string.h>
#define max 30000
long long i,j,n,a[max+1][6],b[6]={0,1,5,10,25,50};
int main()
{
memset(a,0,sizeof(a));
for (i=1;i<6;i++)
{
++a[b[i]][i-1];//加入面值b[I];
for (j=1;j<=max;j++)
{
if (j-b[i]>=0) a[j][i]=a[j][i-1]+a[j-b[i]][i];
else a[j][i]=a[j][i-1];
}
}
while (scanf("%lld",&n)!=EOF)
if (a[n][5]==1||n==0) printf("There is only 1 way to produce %lld cents change.\n",n);//面值爲0也要輸出1種Y-Y
else printf("There are %lld ways to produce %lld cents change.\n",a[n][5],n);
return 0;
}