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+高精度

高精度也可以是把数分成前半部分和后半部分，分别算。注意，要先算前半部分，否则后半部分取完mod再除mod会一直是0。

同时，容易知道(dp1[j]+dp1[j-price[i]])/mod ≤ 1 ，故先算前半部分不用担心错误

提一点，前半部分的计算也是状态的转移。

#include<iostream>
#include<cstdio>
using namespace std;
const long long mod=1e18;
int price[105];
long long dp1[1005],dp2[1005];

int main()
{
    int a,k;
    cin>>a>>k;
    for(int i=1;i<=k;i++) price[i]=i;
    dp1[0]=1;
    for(int i=1;i<=k;i++)
        for(int j=price[i];j<=a;j++)
        {
            dp2[j]=(dp1[j]+dp1[j-price[i]])/mod+dp2[j]+dp2[j-price[i]];
            dp1[j]=(dp1[j]+dp1[j-price[i]])%mod;
        }
    if(dp2[a]) cout<<dp2[a];
    cout<<dp1[a];
    return 0;
}
dp