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Revenge of Fibonacci(杭电5018)
Revenge of Fibonacci
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 721 Accepted Submission(s): 332
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
32 3 52 3 62 2 110
Sample Output
YesNoYes#include<stdio.h>#include<string.h>int s[10000];int main(){ int n,i,a,b,c,k; scanf("%d",&n); while(n--) { memset(s,0,sizeof(s)); scanf("%d%d%d",&a,&b,&c); if(a==c||b==c) //本题容易少考虑的情况。 printf("Yes\n"); else { k=0; s[1]=a,s[2]=b; for(i=3;;) { s[i]=s[i-1]+s[i-2]; if(s[i]<c) { i++; } if(s[i]==c) { k=1; break; } if(s[i]>c) { break; } } if(k==1) printf("Yes\n"); else printf("No\n"); } } return 0;}
Revenge of Fibonacci(杭电5018)
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