洛谷1537（bitset+01背包）

总数是偶数并且其一半可得即可。

bitset的移位可替代原本的数组转移。

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <ctime>
 #include <cctype>
 #include <climits>
 #include <iostream>
 #include <iomanip>
 #include <algorithm>
 #include <string>
 #include <sstream>
 #include <stack>
 #include <queue>
 #include <set>
 #include <map>
 #include <vector>
 #include <list>
 #include <fstream>
 #include <bitset>
 #define init(a, b) memset(a, b, sizeof(a))
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define irep(i, a, b) for (int i = a; i >= b; i--)
 using namespace std;

 typedef double db;
 typedef long long ll;
 typedef unsigned long long ull;
 typedef pair<int, int> P;
 const int inf = 0x3f3f3f3f;
 const ll INF = 1e18;

 template <typename T> void read(T &x) {
     x = ;
     int s = , c = getchar();
     for (; !isdigit(c); c = getchar())
         if (c == '-')    s = -;
     for (; isdigit(c); c = getchar())
         x = x *  + c - ;
     x *= s;
 }

 template <typename T> void write(T x) {
     if (x < )    x = -x, putchar('-');
     if (x > )    write(x / );
     putchar(x %  + '');
 }

 template <typename T> void writeln(T x) {
     write(x);
     puts("");
 }

 const int maxn = 2e4 + ;
 int a[], sum, kase;
 bitset<maxn> bt;

 int main() {
     while () {
         sum = ;
         rep(i, , )    read(a[i]), sum += a[i] * i;
         if (!sum)    break;
         printf("Collection #%d:\n", ++kase);
         if (sum & )    puts("Can't be divided.\n");
         else {
             bt.reset(); bt.set();
             rep(i, , )    rep(j, , a[i])    bt |= bt << i;
             if (bt.test(sum >> ))    puts("Can be divided.\n");
             else    puts("Can't be divided.\n");
         }
     }
     return ;
 }