xtu read problem training A - Dividing

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #define LL long long
13 #define INF 0x3f3f3f3f
14 using namespace std;
15 int dp[200000],w[8] = {0,1,2,3,4,5,6};
16 int num[7],sum,val;
17 int main() {
18     int i,j,k,ks = 1,temp,u;
19     int a[1000];
20     while(true) {
21         for(val = sum = 0,i = 1; i <= 6; i++) {
22             scanf("%d",num+i);
23             sum += num[i];
24             val += i*num[i];
25         }
26         if(!sum) break;
27         printf("Collection #%d:\n",ks++);
28         if(val&1) puts("Can‘t be divided.");
29         else {
30             temp = val>>1;
31             memset(dp,0,sizeof(dp));
32             for(i = 1; i <= 6; i++) {
33                 int u = log2(num[i]);
34                 for(k = 0; k <= u; k++){
35                     if(k == u){a[k] = num[i]-(1<<u)+1;}
36                     else a[k] = 1<<k;
37                 }
38                 for(k = 0; k <= u; k++){
39                     for(j = temp; j >= a[k]*i; j--){
40                         dp[j] = max(dp[j],dp[j-a[k]*i]+a[k]*i);
41                     }
42                 }
43             }
44             if(dp[temp] == temp) puts("Can be divided.");
45             else puts("Can‘t be divided.");
46         }
47         printf("\n");
48     }
49     return 0;
50 }