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 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…
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<<"---"<<endl
typedef 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;
}
