Baby Ming and Weight lifting
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1436 Accepted Submission(s): 524
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 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.
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 minimum a+b)
Sample Input
2
1 2 6
1 4 5
Sample Output
2 2
Impossible
Source
BestCoder Round #69 (div.2)
凑哑铃。。。
#include <iostream> #include <stdio.h> #include <math.h> #include <stdlib.h> #include <string> #include <string.h> #include <algorithm> #include <vector> #include <queue> #include <iomanip> #include <time.h> #include <set> #include <map> #include <stack> using namespace std; typedef long long LL; const int INF=0x7fffffff; const int MAX_N=10000; int T; int a,b,c; int main(){ cin>>T; while(T--){ cin>>a>>b>>c; if(c%2==1){ cout<<"Impossible"<<endl; continue; } int flag=0; if(a>b){ flag=1; swap(a,b); } int ans1,ans2,ss=0; for(int i=0;i<=1000;i++){ int cur=i*a; if(cur>c/2)break; if((c/2-cur)%b==0){ if(flag==1){ ans1=(c/2-cur)/b; ans2=i; } else{ ans1=i; ans2=(c/2-cur)/b; } ss=1; break; } } if(ss==1){ cout<<2*ans1<<" "<<2*ans2<<endl; } else cout<<"Impossible"<<endl; } return 0; }