POJ 3181 Dollar Dayz（完全背包）

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Dollar Dayz

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2275		Accepted: 934

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

Source

USACO 2006 January Silver

题意：求n用1~k来划分的划分数。

思路：注意会爆longlong 要用两个数组来模拟大数，此外就是很简单的完全背包了，数组b来存超过1e18的数字。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
ll a[1005],b[1005];
int n,k;
const long long inf=1e18;
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        a[0]=1;
        for(int i=1;i<=k;i++)
        {
            for(int j=i;j<=n;j++)
            {
                b[j]=b[j]+b[j-i]+(a[j]+a[j-i])/inf;
                a[j]=(a[j]+a[j-i])%inf;
            }
        }
        if(b[n]) printf("%lld",b[n]);
        printf("%lld\n",a[n]);
    }
    return 0;
}