Revenge of Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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 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

Hint

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

水题，fibonacci数不会超过45个就会超范围了，所以最多只有45个有用。

代码：

//0ms
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
long long a[50];
int main()
{
    int t;
    long long c;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d%I64d%I64d",&a[0],&a[1],&c);
        if(a[0]==c||a[1]==c)
        {
            printf("Yes\n");
            continue;
        }
        int sign=0;
        for(int i=2;i<50;i++)
        {
            a[i]=a[i-1]+a[i-2];
            if(a[i]==c)
            {
                sign=1;
                break;
            }
        }
        if(sign)
        printf("Yes\n");
        else
        printf("No\n");
    }
    return 0;
}