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

題意就是很簡單的完全揹包，但是結果特別大。需要用兩個long long 陣列 記錄結果。

陣列 a[] 記錄 高位，陣列b[]記錄低 位 （對 1e18 取模結果）

這樣

a[j] = a[j] + a[j-cost[i]] + (b[j] + b[j-cost[i]])%M

b[j] = (b[j] + b[j-cost[i]])%M  => b[j] = (b[j]%M + b[j-cost[i]]%M)%M;

code : 

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<vector>
#include<map>
#include<algorithm>
#include<string>
const unsigned long long M = 1e18;
using namespace std;
typedef long long LL;
int cost[100+5];
LL a[1000+5];
LL b[1000+5];
int n,k;

int main()
	{
		while(scanf("%d%d",&n,&k) != EOF){
			memset(a,0,sizeof(a));
			memset(b,0,sizeof(b));
			memset(cost,0,sizeof(cost));
//			for(int i = 0; i < k; ++i){
//				cost[i] = i+1;
//			}
			b[0] = 1;
			for(int i = 1; i <= k; ++i){
				for(int j = i; j <= n; ++j){
					a[j] = a[j] + a[j-i] + (b[j] + b[j-i])/M;//記錄高位
					b[j] = (b[j]%M  + b[j-i]%M)%M; //記錄低位
				}
			}			
			if(a[n])
				printf("%I64d",a[n]);
			printf("%I64d\n",b[n]);
		}	
	

		return 0;
	}