HDU 5610 Baby Ming and Weight lifting（枚举）

Baby Ming and Weight lifting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1439    Accepted Submission(s): 525


[align=left]Problem Description[/align]
Baby Ming is fond of weight lifting. He has a barbell pole(the weight of which can be ignored) and two different kinds of barbell disks(the weight of which are respectively
a
and b),
the amount of each one being infinite.

Baby Ming prepare to use this two kinds of barbell disks to make up a new one weighted
C(the
barbell must be balanced), he want to know how to do it.



 

[align=left]Input[/align]
In the first line contains a single positive integer
T,
indicating number of test case.

For each test case:

There are three positive integer a,b,
and C.

1≤T≤1000,0<a,b,C≤1000,a≠b

 

[align=left]Output[/align]
For each test case, if the barbell weighted
C
can’t be made up, print Impossible.

Otherwise, print two numbers to indicating the numbers of
a
and b
barbell disks are needed. (If there are more than one answer, print the answer with minimum
a+b)

 

[align=left]Sample Input[/align]

2
1 2 6
1 4 5

 

[align=left]Sample Output[/align]

2 2
Impossible

 

[align=left]Source[/align]
BestCoder Round #69 (div.2)

 

[align=left]Recommend[/align]
hujie   |   We have carefully selected several similar problems for you:  5674 5673 5672 5671 5669 
把每种情况都枚举一遍。

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int t,m,n,i,j,k,l,sum,minn,maxx,num,flag,a,ans1,ans2;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&m,&n,&sum);
if(sum%2)
{
printf("Impossible\n");
continue;
}
else
{
minn=min(n,m);
maxx=max(n,m);
}
if(maxx*2>sum)//不能用大的
{
if(sum%minn)
printf("Impossible\n");
else
{
if(sum/minn%2)
printf("Impossible\n");
else
{
num=sum/minn;
if(minn==m)
printf("%d 0\n",num);
else
printf("0 %d\n",num);
}
}
}
else
{
flag=0;
l=sum/maxx;
for(i=l;i>=0;i--)
{
if(flag)
break;
if((sum-i*maxx)%minn==0)
{
if(flag)
break;
a=(sum-i*maxx)/minn;
for(j=i;j>=0;j--)
{
if(flag)
break;
for(k=0;k<=a;k++)
{
if(k*minn+j*maxx==sum/2)//如果能构成一半就可以，比较机智的一个地方
{
ans1=i;
ans2=a;
flag=1;
}
}
}
}
}
if(flag==0)
printf("Impossible\n");
if(flag==1)
{
if(minn==m)
printf("%d %d\n",ans2,ans1);
else
printf("%d %d\n",ans1,ans2);
}
}
}
return 0;
}