hdu 5018 Revenge of Fibonacci(BestCoder Round #10)

Revenge of Fibonacci

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

Total Submission(s): 372 Accepted Submission(s): 177



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
HintFor 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;
}