Dollar Dayz---动态规划

​本文已参与「新人创作礼」活动，一起开启掘金创作之路。​
 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

 题意：给出两个数,n,k,问k以内的整数有多少种组成n的方法
这位大佬讲的很详细！可以看看！#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const long long flag=1000000000000000000;///long long型能存19位数1000 0000 0000 0000 000
long long a[1010],b[1010];///高位  低位    1.84*10^19
int main()
{
    int n,k;
    while(~scanf("%d %d",&n,&k))
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        b[0]=1;                    ///低位赋初值
        for(int i=1; i<=k; i++)
        {
            for(int j=i; j<=n; j++)
            {
                a[j]+=a[j-i];
                b[j]+=b[j-i];
                if(b[j]>=flag)
                {
                    a[j]+=b[j]/flag;
                    b[j]%=flag;
                }
            }
        }
        if(a[n])                   ///判断有高位就输出高位
            printf("%lld",a[n]);
        printf("%lld\n",b[n]);
    }
    return 0;
}