1000. Fibonacci

1000. Fibonacci 1

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn-1 + Fn-2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

Given an integer n, your goal is to compute the last Fn mod (10^9 + 7).

Input

The input test file will contain a single line containing n (n ≤ 100).
There are multiple test cases!

Output

For each test case, print the Fn mod (10^9 + 7).

Sample Input

9

Sample Output

34

本题比较简单，由于最多只需要计算到第100个斐波那契序列。所以直接用long long利用for循环方法得到答案。代码如下。

#include <iostream>using namespace std;int main(){    int n;    while(cin>>n)    {        long long f0=0, f1=1;        if(n==0) cout << 0 << endl;        else if(n==1) cout << 1 << endl;        else{            long long f2;            for(int i=2;i<=n;i++)            {                f2=f0+f1;                f0=f1;                f1=f2;            }            f2=f2%(1000000007);            cout << f2 << endl;        }    }    return 0;}             

1001. Fibonacci 2

Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn-1 + Fn-2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
 
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
 
Given an integer n, your goal is to compute the last Fn mod (10^9 + 7).

Input

 The input test file will contain a single line containing n (n ≤ 2^31-1).
There are multiple test cases!
 
 

Output

 For each test case, ple Input
9

Sample Output

34

Hint

 You may need to use “long long”.

#include <cmath>#include <cstring>using namespace std;#define MOD 1000000007struct matrix{    long long a[2][2];};matrix mul(matrix x,matrix y){    matrix m;    memset(m.a,0,sizeof(m.a));    for(int i=0;i<2;i++)        for(int j=0;j<2;j++)            for(int k=0;k<2;k++)                m.a[i][j]=(m.a[i][j]+x.a[i][k]*y.a[k][j])%MOD;    return m;}void power(int n){    matrix t,result;    t.a[0][0]=1;    t.a[0][1]=1;    t.a[1][0]=1;    t.a[1][1]=0;    memset(result.a,0,sizeof(result.a));    for(int i=0;i<2;i++)    result.a[i][i]=1;    while(n)    {        if(n&1) result=mul(result,t);        t=mul(t,t);        n=n>>1;    }    cout << result.a[0][1] << endl;}int main(){    int n;    while(cin>>n)    {        power(n);    }    return 0;}