【HDU

I - Dividing

Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value. 
Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles. 

Input

Each line in the input describes one collection of marbles to be divided. The lines consist of six non-negative integers n1, n2, ..., n6, where ni is the number of marbles of value i. So, the example from above would be described by the input-line ``1 0 1 2 0 0''. The maximum total number of marbles will be 20000. 

The last line of the input file will be ``0 0 0 0 0 0''; do not process this line. 

Output

For each colletcion, output ``Collection #k:'', where k is the number of the test case, and then either ``Can be divided.'' or ``Can't be divided.''. 

Output a blank line after each test case. 

Sample Input

1 0 1 2 0 01 0 0 0 1 10 0 0 0 0 0

Sample Output

Collection #1:Can't be divided.Collection #2:Can be divided.

题意：有价值分别为1~6的大理石，现要将这些大理石分为价值相等的两部分，输出是否可实现分离。

分析：一个裸的混合背包问题。在做背包之前先判断大理石的总价值sum是否为奇数，若为奇数直接判断无法分离。若是偶数则将总价值sum /= 2，以sum值作为背包容量。最后判断dp[sum]是否等于sum，若等于则可分离，否则无法分离。

代码如下：

#include <set>#include <map>#include <queue>#include <cmath>#include <vector>#include <cctype>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define mod 835672545#define INF 0x3f3f3f3f#define LL long longusing namespace std;const int MX = 120005;int n, m;int sum;int a[10];int dp[MX];void zeroOnePack(int v){    for(int i = sum; i >= v; i--){        dp[i] = max(dp[i], dp[i-v] + v);    }}void completePack(int v){    for(int i = v; i <= sum; i++){        dp[i] = max(dp[i], dp[i-v] + v);    }}void mutiPack(int v, int w){    if(v * w >= sum){        completePack(v);        return ;    }    int k = 1;    while(k < w){        zeroOnePack(k*v);        w -= k;        k <<= 1;    }    zeroOnePack(w*v);}int main(){    int cas = 0;    while(~scanf("%d", &a[1])){        memset(dp, 0, sizeof(dp));        sum = a[1];        for(int i = 2; i <= 6; i++){            scanf("%d", &a[i]);            sum += a[i]*i;        }        if(a[1] + a[2] + a[3] + a[4] + a[5] + a[6] == 0)    return 0;        printf("Collection #%d:\n", ++cas);        if(sum & 1){            printf("Can't be divided.\n\n");            continue;        }        sum /= 2;        for(int i = 1; i <= 6; i++){            mutiPack(i, a[i]);        }        if(dp[sum] == sum)   printf("Can be divided.\n\n");        else    printf("Can't be divided.\n\n");    }    return 0;}