poj 3181 Dollar Dayz（完全背包）

Dollar Dayz

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5419		Accepted: 2054

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 Output5

高精度问题，必须分开保存。

 

题意：输入n，和k，问将n用1到k这k个数字进行拆分，有多少种拆分方法。例如：

n=5,k=3 则有n=3+2,n=3+1+1,n=2+1+1+1,n=2+2+1,n=1+1+1+1+1这5种拆分方法

 

附上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
    __int64 a[1100],b[1100],inf;
    int n,k,i,j;
    inf=1;
    for(i=0; i<18; i++)
        inf*=10;
    while(~scanf("%d%d",&n,&k))
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        a[0]=1;
        for(i=1; i<=k; i++)
            for(j=1; j<=n; j++)
            {
                if(j<i) continue;
                b[j]=b[j]+b[j-i]+(a[j]+a[j-i])/inf;  //高精度处理
                a[j]=(a[j]+a[j-i])%inf;
            }
        if(b[n]) printf("%I64d",b[n]);
        printf("%I64d\n",a[n]);
    }
    return 0;
}