"Cattle-off practice match 53A" Beyond sister love string

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Portal

Portal1: Nowcoder

Description

Beyond sister really likes her name, that she only likes the letters \ (\ textrm { "c" } \) and \ (\ {textrm "the y-"} \) . Thus contains only beyond sister like \ (\ textrm { "c" } \) and \ (\ textrm { "y" } \) of the string, and the string can not be two consecutive \ (\ textrm { " C "} \) . Please find how many of length \ (n \) strings are beyond the sister school like string. The answer to \ (1e9 + 7 \) modulo.

Input

Enter an integer \ (n-\) .

\ (1 \ n \ the 100000 \) .

Output

Output a integer answer.

Sample Input

3

Sample Output

5

Sample Explain

\(\textrm{cyy, cyc, yyy, yyc, ycy}\)。

Solution

We can see by enumerating

When \ (n = 1 \) , the answer is \ (2 \) : c, y;

When \ (n = 2 \) , the answer is \ (3 \) : cy, yc, yy;

When \ (n = 3 \) , the answer is \ (5 \) : cyy, cyc, yyy, yyc, ycy;

When \ (n = 4 \) , the answer is \ (8 \) : yyyy, yyyc, yycy, ycyy, cyyy, cycy, yccy, ycyc;

When \ (n = 5 \) , the answer is \ (13 \) : yyyyy, yyyyc, yyycy, yycyy, ycyyy, cyyyy, yycyc, ycyyc, cyyyc, ycycy, cyycy, cycyy, cycyc;

\(\cdots \cdots\)

Easily summed law: \ (\ TEXTRM {F (I) = F (I -. 1) + F (I - 2)} (X \ GE. 3) \)

When finished the code required for the \ (n = 1 \) Patent judgment time.

Code

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

typedef long long LL;
const int mod = 1e9 + 7;
int n;
int main() {
    scanf("%d", &n);
    if (n == 1) {//特判n = 1的情况
        printf("2\n");
        return 0;
    }
    LL x1 = 2, x2 = 3;
    for (int i = 3; i <= n; i++) {
        LL tmp = x1;
        x1 = x2 % mod;
        x2 = (x2 + tmp) % mod;//前两项的和
    }
    printf("%lld\n", x2 % mod);//不要忘记取余
    return 0;
}