完全背包+高精度 poj3181

                                                                                           Dollar Dayz

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2498		Accepted: 1022

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：
已知n和k，求用小于等于k的数组成n的方案数。
解题思路：可以采用dp递推的思路，也可以开一个一维数组，完全背包背一遍，由于最后结果最大可以达到10^33，超出long long的数据范围，因此需要用两个long long来保存结果，
完全背包1A，爽。。。
完全背包代码：
#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
typedef __int64 ll;
struct node
{
       ll a,b;
       bool flag;
};
const ll mod=100000000000000000ll;
node dp[1110];
int main()
{
       ll i,j,k,m,n,p,q;
       while(~scanf("%I64d%I64d",&n,&k))
       {
              for(i=0;i<=1100;i++)
              {
                  dp[i].a=0;
                  dp[i].b=0;
                  dp[i].flag=0;
              }
              dp[0].flag=1;
              dp[0].b=1;
              for(i=1;i<=k;i++)
                 for(j=0;j<=n;j++)
                 if(dp[j].flag)
                 {
                        dp[j+i].flag=1;
                        p=dp[j+i].b+dp[j].b;
                        dp[j+i].b=p%mod;
                        dp[j+i].a=dp[j+i].a+dp[j].a+p/mod;
                 }
                 if(dp[n].a==0)printf("%I64d\n",dp[n].b);
                 else
                 {
                        printf("%I64d",dp[n].a);
                        printf("%017I64d\n",dp[n].b);
                 }
       }
       return 0;
}