POJ - 3181

Farmer John goes to Dollar Days at The Cow Store and discovers an unlimited number of tools on sale. During his first visit, the tools are selling variously for $1, $2, and $3. Farmer John has exactly $5 to spend. He can buy 5 tools at $1 each or 1 tool at $3 and an additional 1 tool at $2. Of course, there are other combinations for a total of 5 different ways FJ can spend all his money on tools. Here they are:

        1 @ US$3 + 1 @ US$2

        1 @ US$3 + 2 @ US$1

        1 @ US$2 + 3 @ US$1

        2 @ US$2 + 1 @ US$1

        5 @ US$1

Write a program than will compute the number of ways FJ can spend N dollars (1 <= N <= 1000) at The Cow Store for tools on sale with a cost of $1..$K (1 <= K <= 100).

Input

A single line with two space-separated integers: N and K.

Output

A single line with a single integer that is the number of unique ways FJ can spend his money.

Sample Input

5 3

Sample Output

5

完全背包+大数:

#include<stdio.h>
int a[1010][105];
int dsj(int n,int m)
{
    int carry=0;
    for(int i=0;i<=100;i++)
    {
        a[m][i]=a[m][i]+a[n][i]+carry;
        carry=a[m][i]/10000;
        a[m][i]%=10000;
    }
}
int putout(int n)
{
    int i=100;
    for(;i>=0;i--)
    {
        if(a[n][i])
            break;
    }
    printf("%d",a[n][i]);
    i--;
    for(;i>=0;i--)
    {
        printf("%04d",a[n][i]);
    }
    printf("\n");
}
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    a[0][0]=1;
    for(int i=1;i<=k;i++)
    {
        for(int j=0;j<=n-i;j++)
        {
            if(a[j])
            {
                dsj(j,j+i);//a[j+i]加上a[j]的值
            }
        }
    }
    putout(n);
}