POJ 3181 - Dollar Dayz（完全背包+高精度）

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：给出总钱数 n 和钱币种类 k 种 ，钱币数量不限，求出能组成钱数的方案数。

思路：完全背包，前面的方案数应该与后面的方案数叠加，所以状态转移方程是 dp[j]+=dp[j-i]；

           小数据测试正确之后交上去WA了，在测试 1000 100这组数据的时候爆了，使用高精度的思想，定义两个 long long 的数组，一个记录高位，一个记录低位。

CODE：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
const int inf=0xfffffff;
typedef long long ll;
using namespace std;

ll dp1[1005],dp2[1005];
int main()
{
int n,k;
ll mod=1;
for(int i=1;i<=18;i++) mod*=10;
while(~scanf("%d %d",&n,&k)){
memset(dp1,0,sizeof(dp1));
memset(dp2,0,sizeof(dp2));
dp2[0]=1;
int ans=0;
for(int i=1;i<=k;i++){
for(int j=i;j<=n;j++){
if(dp2[j-i]){
dp1[j]+=(dp1[j-i]+(dp2[j-i]+dp2[j])/mod);
dp2[j]=(dp2[j-i]+dp2[j])%mod;
}
}
}
if(dp1
)printf("%I64d",dp1
);
printf("%I64d\n",dp2
);
}
return 0;
}