hdoj 1898 Sempr == The Best Problem Solver?

Sempr == The Best Problem Solver?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1490    Accepted Submission(s): 970

Problem Description

As is known to all, Sempr(Liangjing Wang) had solved more than 1400 problems on POJ, but nobody know the days and nights he had spent on solving problems.
Xiangsanzi(Chen Zhou) was a perfect problem solver too. Now this is a story about them happened two years ago.
On March 2006, Sempr & Xiangsanzi were new comers of hustacm team and both of them want to be "The Best New Comers of March", so they spent days and nights solving problems on POJ.
Now the problem is below: Both of them are perfect problem solvers and they had the same speed, that is to say Sempr can solve the same amount of problems as Xiangsanzi, but Sempr enjoyed submitting all the problems at the end of every A hours but Xiangsanzi enjoyed submitting them at the end of every B hours. In these days, static(Xiaojun Wu) was the assistant coach of hustacm, and he would check the number of problems they solved at time T. Give you three integers A,B,and T, you should tell me who is "The Best New Comers of March". If they solved the same amount of problems, output "Both!". If Sempr or Xiangsanzi submitted at time T, static would wait them.

 

 

Input

In the first line there is an integer N, which means the number of cases in the data file, followed by N lines.
For each line, there are 3 integers: A, B, T.
Be sure that A,B and N are no more than 10000 and T is no more than 100000000.

 

 

Output

For each case of the input, you should output the answer for one line. If Sempr won, output "Sempr!". If Xiangsanzi won, output "Xiangsanzi!". And if both of them won, output "Both!".

 

 

Sample Input

3

2 3 4

2 3 6

2 3 9

 

 

Sample Output

Sempr!

Both!

Xiangsanzi!

 

#include<stdio.h>
int main()
{
	int n,a,b,t;
	int suma,sumb;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%d%d%d",&a,&b,&t);
		suma=a*(t/a);
		sumb=b*(t/b);
		if(suma==sumb)
		printf("Both!\n");
		else if(suma<sumb)
		printf("Xiangsanzi!\n");
		else if(suma>sumb)
		printf("Sempr!\n");
	}
	return 0;
}