Revenge of Fibonacci
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 797 Accepted Submission(s): 360
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.
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
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
/*
HDU 5018 斐波拉契数
给定初始的两个数 判断第三个数是否在是该数列
*/
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
int t;
__int64 a,b,c,sum;
scanf("%d",&t);
while(t--)
{
scanf("%I64d%I64d%I64d",&a,&b,&c);
sum=a+b;
if(a==c||b==c)
{
printf("Yes\n");
continue;
}
while(sum<c)
{
a=b;
b=sum;
sum=a+b;
}
if(sum==c)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}