packets

packets  

时间限制(普通/Java):1000MS/10000MS     运行内存限制:65536KByte
总提交: 27            测试通过: 14

描述

A factory produces products packed in square packets of the same height h and of the sizes 1*1, 2*2, 3*3, 4*4, 5*5, 6*6. These products are always delivered to customers in the square parcels of the same height h as the products have and of the size 6*6. Because of the expenses it is the interest of the factory as well as of the customer to minimize the number of parcels necessary to deliver the ordered products from the factory to the customer. A good program solving the problem of finding the minimal number of parcels necessary to deliver the given products according to an order would save a lot of money. You are asked to make such a program.

输入

The input file consists of several lines specifying orders. Each line specifies one order. Orders are described by six integers separated by one space representing successively the number of packets of individual size from the smallest size 1*1 to the biggest size 6*6. The end of the input file is indicated by the line containing six zeros.

输出

The output file contains one line for each line in the input file. This line contains the minimal number of parcels into which the order from the corresponding line of the input file can be packed. There is no line in the output file corresponding to the last ``null'' line of the input file.

样例输入

0 0 4 0 0 1
7 5 1 0 0 0
0 0 0 0 0 0 

样例输出

2
1

题目上传者

crq

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
 
using namespace std;
int v[10];
int ava[10];
int cnt = 0;
 
void work() {
    cnt = 0;//init
    int res;//剩余的
    cnt += v[6];//v[6] == 0;
    if(v[5]) { //v[5] == 0;
        cnt += v[5];
        ava[1] += v[5]*11;//ava[1] += 11;!!!
    }
    if(v[4]) {
        cnt += v[4];
        ava[2] += v[4] * 5;
    }
    if(v[3]) {
        int tag = true;
        cnt += v[3]/4 ;//v[3] < 4; 3*3*4 == 36;
        res = v[3]%4; //占用了几个.
 
        if(res==0) tag = false;
        else cnt++;
 
        if(res==1 && tag) {//占用一个3*3
            ava[2] += 5;
            ava[1] += 7;
            tag = false;
        }
        if(res==2 && tag){//two 3*3
            ava[2] += 3;
            ava[1] += 6;
            tag = false;
        }
        if(res==3 && tag) {
            ava[2] += 1;
            ava[1] += 5;
            tag = false;
        }
    }
    if(v[2]) {
        if(ava[2] >= v[2]) {
            ava[1] += (ava[2]-v[2]) * 4;
        }
        else {
            v[2] -= ava[2];
            cnt += v[2]/9;//2*2*9 = 36
            res = v[2]%9;
            if(res != 0) cnt++;
            ava[1] += (9-res) * 4;
        }
    }
    if(v[1]) {
        if(ava[1]>=v[1]) {
            return;
        }
        else {
            v[1] -= ava[1];
            cnt += v[1]/36;
            res = v[1]%36;
            if(res) cnt++;
        }
    }
    return;
}
 
int main()
{
    memset(ava, 0, sizeof(ava));
    memset(v, 0, sizeof(v));
    while(scanf("%d%d%d%d%d%d", &v[1], &v[2], &v[3], &v[4], &v[5], &v[6])==6) {
        if(v[0]==v[1] && v[1]==v[2]&&v[2]==v[3]&&v[3]==v[4]&&v[4]==v[5]&&
           v[5]==v[6]&&v[6]==0) break;
        work();
        printf("%d\n", cnt);
        memset(ava, 0, sizeof(ava));
        memset(v, 0, sizeof(v));
    }
    return 0;
}