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杭电5018--Revenge of Fibonacci

时间:2015-06-15 00:05:48      阅读:103      评论:0      收藏:0      [点我收藏+]

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Revenge of Fibonacci

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 6   Accepted Submission(s) : 2

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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

 

Hint

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

Source

BestCoder Round #10 
 
 1 #include<stdio.h>
 2 long long set[10000100];
 3 int main()
 4 {
 5     int t;
 6     scanf("%d",&t);
 7     while(t--)
 8     {
 9         long long a,b,c,i;
10         scanf("%lld %lld %lld",&a,&b,&c);
11         set[0]=a; set[1]=b;
12         if(a==c||b==c) //无数次栽在初始条件判断上; 
13         printf("Yes\n");
14        else
15         {
16           for(i=2;;i++)
17             {
18                 set[i]=set[i-1]+set[i-2];
19                 if(set[i]==c)
20                 {
21                     printf("Yes\n");
22                     break;
23                 }
24                 if(set[i]>c)
25                 {
26                     printf("No\n");
27                     break; 
28                 }
29             }
30         }
31     }
32     return 0;
33 } 

 

 

杭电5018--Revenge of Fibonacci

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原文地址:http://www.cnblogs.com/fengshun/p/4575907.html

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