【CUGBACM15级BC第10场 A】hdu 5018 Revenge of Fibonacci

Revenge of Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1288    Accepted Submission(s): 605

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

32 3 52 3 62 2 110

 

Sample Output

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

 

题意：ab为斐波那契数列前两项，问你c 是不是斐波那契数列里面的数

思路：水题，直接模拟即可

#include <iostream>#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <algorithm>#include <functional>#define PI acos(-1)#define eps 1e-8#define inf 0x3f3f3f3f#define debug(x) cout<<"---"<<x<<"---"<<endltypedef long long ll;using namespace std;int main(){    int t;    ll a, b, c;    scanf("%d", &t);    while (t--)    {        scanf("%lld %lld %lld", &a, &b, &c);        if (c == a || c == b)        {            printf("Yes\n");        }        else if (c < a)        {            printf("No\n");        }        else        {            ll xx = a;            int flag = 0;            while (xx <= c)            {                if (xx == c)                {                    flag = 1;                }                xx = a + b;                a = b;                b = xx;            }            if (flag == 1)            {                printf("Yes\n");            }            else            {                printf("No\n");            }        }    }    return 0;}