HDU 1059 Dividing 多重背包

题意:有$n1,n2,n3,n4,n5,n6$个质量分别为$1,2,3,4,5,6$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$dp[n/2]$是否能达到即可。

http://acm.hdu.edu.cn/showproblem.php?pid=1059

/*********************************************
    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[N]) 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;
}