Packets
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 43189 | Accepted: 14550 |
Description
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.
Input
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.
Output
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.
Sample Input
0 0 4 0 0 1
7 5 1 0 0 0
0 0 0 0 0 0
Sample Output
2
1 题意:工厂的包装盒都是6*6规格的，但是产品有1*1、2*2、3*3、4*4、5*5、6*6规格的，输入给出每种产品的个数，输出最少使用的包装盒数。
思路：由大到小装入，按实际剩余空间用2*2或1*1的填充。
AC代码：
#include<iostream>
#include<cstdio>
using namespace std;
int b[], cot;
void test()
{
for(int i = ; i < ; i++)
cout<<b[i]<<" ";
cout<<cot<<endl;
}
int main()
{
freopen("1017.in","r",stdin);
freopen("1017.out", "w", stdout);
while(){
for(int i = ; i < ; i++){
scanf("%d", &b[i]);
}
int flag = ;
for(int i = ; i < ; i++){
if(b[i] == ) flag ++;
}
if(flag == ) break; cot = ;
if(b[] != ){
cot+=b[];
}
// test();
if(b[] != ){
cot+=b[];
int x = b[]*;
b[] = b[]-x>=?b[]-x:;
}
//test();
if(b[] != ){
int res = ;
cot+=b[];
res = *b[];
if(b[] != ) {
int x = b[] / ;
if( x >= b[]) { b[] = b[] - b[]*; res = ; }
else { res -= b[]*; b[] = ; }
}
if(b[] != && res != ) {
b[] = b[]>res?b[]-res:;
}
}
// test();
if(b[] != ) {
int x = b[] / , y = b[] % ;
int res = ;
cot += x;
if(y != ) cot++;
if(y == ){
res = b[] > ?:-b[]*;
b[] = b[] > ? b[]-:;
}
if(y == ){
res = b[]>? :-b[]*;
b[] = b[] > ? b[]-:;
}
if(y == ){
if(b[] != ) { res = ; b[] --; }
else res = ;
}
//cout<<"*"<<res<<endl;
if(res != &&b[] != ) b[] = b[]>res?b[]-res:;
}
// test();
if(b[] != ){
int y = b[] % ;
cot += b[]/;
if(y != ) {
cot++;
y = - y*;
b[] = b[]>y?b[]-y:;
}
}
// test();
if(b[] != ) {
cot += (b[] / );
b[] %= ;
if(b[] != ) cot++;
}
// test();
printf("%d\n", cot);
}
return ;
}
总结：贪心地做要比较细心，WA了很多次都是因为代码里的一点小问题，不用多组数据测都没发现。
下面是1000组数据的输入输出：
View data input:
View data output