POJ 3181（完全背包、高精度）

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/Tianweidadada/article/details/83004555

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9693		Accepted: 3578

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

题意就是很简单的完全背包，但是结果特别大。需要用两个long long 数组 记录结果。

数组 a[] 记录 高位，数组b[]记录低 位 （对 1e18 取模结果）

这样

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;//记录高位
					b[j] = (b[j]%M  + b[j-i]%M)%M; //记录低位
				}
			}			
			if(a[n])
				printf("%I64d",a[n]);
			printf("%I64d\n",b[n]);
		}	
	

		return 0;
	}