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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> Revenge of Fibonacci(杭電5018)

Revenge of Fibonacci(杭電5018)

編輯:C++入門知識

Revenge of Fibonacci(杭電5018)


Revenge of Fibonacci

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 721 Accepted Submission(s): 332


Problem Description In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation
Fn = Fn-1 + Fn-2
with seed values F1 = 1; F2 = 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
#include
#include
int s[10000];
int main()
{
    int n,i,a,b,c,k;
	scanf("%d",&n);
	while(n--)
	{
		memset(s,0,sizeof(s));
	    scanf("%d%d%d",&a,&b,&c);
		if(a==c||b==c)  //本題容易少考慮的情況。
		printf("Yes\n");
		else
		{
			k=0;
		    s[1]=a,s[2]=b;
			for(i=3;;)
			{
			    s[i]=s[i-1]+s[i-2];
				if(s[i]c)
				{
				   break;
				}
			}
			if(k==1)
			printf("Yes\n");
			else
			printf("No\n");
		}
	}
	return 0;
}

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