hdoj 5018 Revenge of Fibonacci

http://acm.hdu.edu.cn/showproblem.php?pid=5018

Revenge of Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 724    Accepted Submission(s): 335

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

   

    

     Hint
    
For the third test case, the new Fibonacci sequence is: 2, 2, 4, 6, 10, 16, 26, 42, 68, 110…
   
  

 

大体思路就是一个一个数判断，因为C<=1000000000,所以是可行的，开始写先把所有的斐波那契数存到数组中，那样超内存了，后来又改为直接比较。

#include<stdio.h>
#include<string.h>
#define MAX 1000000000
__int64 a,b,C,c;
int main()
{
	int N,i,j,k,J;
	scanf("%d",&N);
	while(N--)
    {
    	scanf("%I64d %I64d %I64d",&a,&b,&C);
    	
    	if(C==a||C==b)
    	{
		    printf("Yes\n");
		    continue;
		}
        for(i=3,J=0,c=0;c<=MAX+10;i++)
        {
		    c=a+b;
        	//printf("%I64d %I64d %I64d#\n",a,b,c);
		   
		   a=b;
		   b=c;
		   if(c==C)
		   {
		   	  J=1;
		   	  break;
		   }
	    }
	    if(J==1)
	    printf("Yes\n");
	    else
	    printf("No\n");
	    
	}
	return 0;
}