HDU 1059 Dividing 多重背包

题意:有n1,n2,n3,n4,n5,n6n1,n2,n3,n4,n5,n6个质量分别为1,2,3,4,5,61,2,3,4,5,6的大理石。问能不能把它们分成相同的两份?

思路：多重背包。求出dp[n/2]dp[n/2]是否能达到即可。

/*********************************************
Problem : HDU 1059
Author  : NMfloat
InkTime (c) NM . All Rights Reserved .
********************************************/

#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++)
#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --)
#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)
#define cls(a,x)   memset(a,x,sizeof(a))
#define eps 1e-8

using namespace std;

const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5+5;
const int MAXE = 2e5+5;

typedef long long LL;
typedef unsigned long long ULL;

int T,n,m,k;

int fx[] = {0,1,-1,0,0};
int fy[] = {0,0,0,-1,1};

int dp[500005];
int num[7];
int N;
int ok = 0;
int base[32];

void ZEROONEPACK(int value) {
rrep(i,value,N) {
if(dp[i-value]) dp[i] = 1;
}
}

void COMPPACK(int value) {
rep(i,value,N) {
if(dp[i-value]) dp[i] = 1;
}
}

void input() {

}

void solve() {
if(N&1) return ;
cls(dp,0);
dp[0] = 1 ; N = N / 2;
rep(i,1,6) {
if(i*num[i] >= N) COMPPACK(i);
else {
int tmp = num[i];
int pos = 1;
while(tmp > base[pos]) {
ZEROONEPACK(base[pos]*i);
tmp -= base[pos];
pos ++;
}
if(tmp) ZEROONEPACK(tmp*i);
}
}
if(dp
) ok = 1;
}

int main(void) {
//freopen("a.in","r",stdin);
base[1] = 1;
rep(i,2,30) base[i] = base[i-1] * 2;
int CASENUM = 0;
while(1) {
N = 0 ; ok = 0;
rep(i,1,6) { scanf("%d",&num[i]) ; N += i * num[i];}
if(N == 0) break;
input();
solve();
printf("Collection #%d:\n",++CASENUM);
if(ok) puts("Can be divided.\n");
else puts("Can't be divided.\n");
}
return 0;
}