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HDU 5018 Revenge of Fibonacci(数学)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5018
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 YesHintFor the third test case, the new Fibonacci sequence is: 2, 2, 4, 6, 10, 16, 26, 42, 68, 110…
Source
BestCoder Round #10
代码如下:
#include<cstdio> #include<cstring> using namespace std; typedef __int64 LL; int main() { LL f[2017]; int t; LL a, b, c; scanf("%d",&t); while(t--) { scanf("%I64d %I64d %I64d",&a,&b,&c); f[0]=a, f[1]=b; LL i=2; int flag = 0; if(a==c || b==c) { printf("Yes\n"); continue; } while(1) { f[i]=f[i-1]+f[i-2]; if(f[i] == c) { flag = 1; break; } if(f[i] > c) break; i++; } if(flag) printf("Yes\n"); else printf("No\n"); } return 0; }
HDU 5018 Revenge of Fibonacci(数学)
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