連結：https://codeforces.com/gym/102900/problem/G

G - Fibonacci

In mathematics, the Fibonacci numbers, commonly denoted asfnfn, is a sequence such that each number is the sum of the two preceding numbers, starting with1and1. That is,f1=1, f2=1andfn=fn−2+fn−1(n≥3).

Thus, the beginning of the sequence is1,1,2,3,5,8,13, 21, …

Givenn, please calculate\(\sum_{i=1}^{n}\sum_{j=i+1}^{n}g(fi,fj)\), where\(g(x,y)=1\)when\(x\times y\)is even, otherwise\(g(x,y)=0\).

Input

The only line contains one integern(1≤n≤1e9).

Output

Output one number –\(\sum_{i=1}^{n}\sum_{j=i+1}^{n}g(fi,fj)\)

題目大意：

求長度為n的斐波那契數列中滿足x < y, 且x × y為偶數的點對的數目。

題目分析：

x × y為偶數當且僅當x和y不同時為奇數。本題只需要統計斐波那契數列中奇數和偶數的個數，用所有點對數目減去奇數點對的數目即可。

根據斐波那契數列的規律，數列中元素的奇偶性為：奇，奇，偶，奇，奇，偶……則長度為n的斐波那契數列中偶數的個數為 n / 3。

因此，所求滿足條件的點對數目為：\(\textrm{C}_{n}^{2}-\textrm{C}_{n-n/3}^{2}\)

程式碼實現：

#include <bits/stdc++.h>
using namespace std;
int n;
typedef long long ll;
int k;
ll ans;
int
 
 main(){
    scanf("%d",&n);
    k=n / 3;
    ans=(2*n*k-k*k-k) / 2;
    printf("%lld",ans);
    return 0;
}