ICPC训练联盟2021寒假冬令营 Dollar Dayz

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;
}