Dollar Dayz（大数母函数，高低位存取）

Dollar Dayz

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5655		Accepted: 2125

Description

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
题解：母函数的裸题；但是一交就Wa；最终看了discuss，原来是大数；又不是太大，高低位存下，然而还是Wa，因为MOD还是开小了，MOD开到10^18次方才A；协会大神真是吊，这题
我不看题解肯定想不到会是大数。。。
代码：

extern "C++"{
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
const int INF = 0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
typedef long long LL;
typedef unsigned long long ULL;

void SI(int &x){scanf("%d",&x);}
void SI(double &x){scanf("%lf",&x);}
void SI(LL &x){scanf("%lld",&x);}
void SI(char *x){scanf("%s",x);}
#define MOD 1000000000000000000 
}
const int MAXN = 10010;
struct Node{
    LL h,l;
};
Node a[MAXN],b[MAXN];
int main(){
    int n,k;
    while(~scanf("%d%d",&n,&k)){
        mem(a,0);mem(b,0);
        for(int i = 0;i <= n;i++)a[i].l = 1;
        for(int i = 2;i <= k;i++){
            for(int j = 0;j <= n;j++){
                for(int t = 0;j + t <= n;t += i){
                    b[j + t].l += a[j].l;
                    b[j + t].h += a[j].h;
                    b[j + t].h += b[j + t].l / MOD;
                    b[j + t].l %= MOD;
                }
            }
            for(int j = 0;j <= n;j++){
                a[j].h = b[j].h;
                a[j].l = b[j].l;
                b[j].h = 0;
                b[j].l = 0;
            }
        }
        if(a[n].h)
            printf("%lld",a[n].h);
        printf("%lld\n",a[n].l);
    }
    return 0;
}