poj3181 完全背包+整数拆分

Dollar Dayz

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7840 Accepted: 2922

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

一开始没看出来是完全背包。第一次写了交上去WA，百度了才知道需要用到高精度。

啊日常哀叹自己这个菜鸟。。。（悲恸）

#include<iostream>#include<cstring>#include<iomanip>using namespace std;int n,k;long long  hi[1002],lo[1002],mod = 1;int main(){memset(lo,0,sizeof(lo));memset(hi,0,sizeof(hi));for (int i=0; i<18; i++) mod *= 10;scanf("%d%d",&n,&k);lo[0] = 1;for(int i = 1;i<=k;i++){for(int j = i;j<=n;j++){long long tp = lo[j-i]+lo[j];hi[j] += hi[j-i] + tp / mod;lo[j] = tp % mod;}}if(hi[n]){cout<<hi[n]<<setw(18)<<setfill('0')<<lo[n]<<endl;}else{cout<<lo[n]<<endl;}return 0;}