hdu-5018

Revenge of Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 703    Accepted Submission(s): 327

Problem Description

In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation
F _n = F _n-1 + F _n-2
with seed values F ₁ = 1; F ₂ = 1 (sequence A000045 in OEIS).
---Wikipedia

Today, Fibonacci takes revenge on you. Now the first two elements of Fibonacci sequence has been redefined as A and B. You have to check if C is in the new Fibonacci sequence.

 

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case only contains three integers A, B and C.

[Technical Specification]
1. 1 <= T <= 100
2. 1 <= A, B, C <= 1 000 000 000

 

Output

For each test case, output “Yes” if C is in the new Fibonacci sequence, otherwise “No”.

 

Sample Input

  

   3
2 3 5
2 3 6
2 2 110
  

 

Sample Output

  

   Yes
No
Yes
   
题目分析：这道题目刷的我都快哭了，一直不通过，一看别人的解释我才恍然大悟，尼玛的我竟然把a,b本身给忘了，这道题目错的值。。。
让我伤心了这么久。。。
代码附上：
#include<stdio.h>
int main(){
	int T;
	__int64 f1,f2,f3;
	__int64 a,b,c;
	int i;
	scanf("%d",&T);
	while(T--){
		int flag=0;
		scanf("%I64d%I64d%I64d",&a,&b,&c);
		f1=a;
		f2=b;
		if(f1==c||f2==c){
			printf("Yes\n");       //需要判断C是否是A或B，这代题目真把我给伤的不轻。。。。 
			continue;
		}
		for(;;i++){
			f3=f1+f2;
			f1=f2;
			f2=f3;
			if(f3==c){
				flag=1;
				break;
			}
		   if(f3>c)
		      break;
		}
		if(flag==1)
		    printf("Yes\n");
		else
		    printf("No\n");
	}
	return 0;
}