Even Fibonacci Numbers
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be: 1,2,3,5,8,13,21,3,55,89,...
By considering the terms in the Fibonacci sequence whose values do not exceed N, find the sum of the even-valued terms.
Input Format
First line contains T that denotes the number of test cases. This is followed by T lines, each containing an integer,N.
Constraints
1<=T<=10^5
10<=N<=4*10^16
Output Format
Print the required answer for each test case.
Sample Input 0
2
10
100
Sample Output 0
10
44
Explanation 0
For N=10, we have {2,8}, sum is 10.
For N=100, we have {2,8,34}, sum is 44.
Solution:
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main(){
int t;
scanf("%d",&t);
for(int a0 = 0; a0 < t;
long n,sum=2,p1 = 1,
scanf("%ld",&n);
fib= p1 + p2;
while (fib <= n) {
if(fib%2==0)
sum=sum+fib;
p1 = p2;
p2 = fib;
fib = p1 + p2;
}
printf("%ld\n",sum);
}
return 0;
}
You can also run it on an online IDE:
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