poj-3181 Dollar Dayz

完全背包。+实现俩个大数相加。

http://poj.org/problem?id=3181

Dollar Dayz

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3431		Accepted: 1356

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

#include<iostream>
#include<cstring>
using namespace std;
__int64 temp,a[1010],b[1010];
int main()
{
    int i,j,n,k;
     cin>>n>>k;
      temp=1;
      memset(a,0,sizeof(a));
      memset(b,0,sizeof(b));
     for(i=0;i<18;i++)
       temp*=10;//__int64存放18位。
     a[0]=1;
     for(i=1;i<=k;i++)//i是金币
      for(j=1;j<=n;j++)//j是总额。
         {
             if(i>j)
               continue; 
               b[j]=b[j]+b[j-i]+(a[j]+a[j-i])/temp;//加上进位，以18位为一个进位。
            a[j]=(a[j]+a[j-i])%temp;//不能调换
           
             
         }
         if(b[n])
          cout<<b[n];
          cout<<a[n]<<endl;
   return 0;
}