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

Problem Description

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 respectivelya 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 weightedC(the barbell must be balanced), he want to know how to do it.

 

Input

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

 

Output

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 minimuma+b)

 

Sample Input

21 2 61 4 5 

 

Sample Output

2 2Impossible

 

Source

BestCoder Round #69 (div.2)

 

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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;}