hdu 1059 Dividing

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <queue>
 4 #include <iostream>
 5 #include <cmath>
 6 #include <algorithm>
 7 using namespace std ;
 8 int dp[65000] ;
 9 int r[10] ;
10 int sum, num ;
11 int c = 1 ;
12 void _zeroone(int c, int w)//0-1背包处理单件物品
13 {
14     for(int i = sum; i >= c; --i)
15         dp[i] = max(dp[i], dp[i - c] + w) ;
16 }
17 void _compl(int c, int w)//完全背包处理单件物品
18 {
19     for(int i = c; i <= sum; ++i)
20         dp[i] = max(dp[i], dp[i - c] + w) ;
21 }
22 void  _mult(int c, int w, int amount)
23 {
24     if(c * amount >= sum) { _compl(c, w) ; return ; }//满足条件时,该件物品可以直接当完全背包时的单件物品处理
25     int k = 1 ;
26     while(k < amount)//二进制分解
27     {
28         _zeroone(c * k,w * k) ;
29         amount -= k ;
30         k = k * 2 ;
31     }
32     _zeroone(amount * c, amount * w) ;
33 }
34 int main()
35 {
36     while(1)
37     {
38         memset(dp, 0, sizeof dp) ;
39         sum = 0 ;
40         int flag = 0 ;
41         for(int i = 1; i <= 6; ++i){
42             scanf("%d",&r[i]) ;
43             if(r[i] != 0) flag = 1 ;
44             sum += i * r[i] ;
45         }
46         if(!flag) break ;
47         printf("Collection #%d:\n",c++) ;
48         if(sum & 1) { printf("Can‘t be divided.\n\n") ; continue ;}
49         sum /= 2 ;
50         for(int i = 1; i <= 6; ++i)//将多重背包每件分开转换为完全背包或0-1背包中的单件物品处理
51         _mult(i,i,r[i]) ;
52         if(dp[sum] == sum ) printf("Can be divided.\n") ;
53         else printf("Can‘t be divided.\n") ;
54         printf("\n") ;
55     }
56     return 0 ;
57 }