cf 450b 矩阵快速幂（数论取模 一大坑点啊）

Posted 2020-10-15

Jzzhu has invented a kind of sequences, they meet the following property:

You are given x and y, please calculate f_n modulo 1000000007 (10⁹?+?7).

Input

The first line contains two integers x and y (|x|,?|y|?≤?10⁹). The second line contains a single integer n (1?≤?n?≤?2·10⁹).

Output

Output a single integer representing f_n modulo 1000000007 (10⁹?+?7).

Example

Input

2 3
3

Output

1

Input

0 -1
2

Output

1000000006

Note

In the first sample, f₂?=?f₁?+?f₃, 3?=?2?+?f₃, f₃?=?1.

In the second sample, f₂?=??-?1; ?-?1 modulo (10⁹?+?7) equals (10⁹?+?6).

 

题意 ： 显然是矩阵快速幂么

坑点 ： 就是数论取模这一块 ， 稍不注意就错了

const ll mod = 1e9+7;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
#define Max(a,b) a>b?a:b
#define Min(a,b) a>b?b:a


struct mat
{
    ll a[2][2];
};

mat mul(mat a, mat b){
    mat r;
    memset(r.a, 0, sizeof(r.a));
    
    for(int i = 0; i < 2; i++){
        for(int k = 0; k < 2; k++){
            if (a.a[i][k]){
                for(int j = 0; j < 2; j++){
                    if (b.a[k][j]){
                        r.a[i][j] += (a.a[i][k] * b.a[k][j] + mod)%mod;
                        r.a[i][j] = (r.a[i][j] + mod)%mod;
                    }
                }
            }
        }
    }
    return r;
}

mat pow(mat a, int n){
    mat b;
    
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++)
            if (i == j) b.a[i][i] = 1;
            else b.a[i][j] = 0;
   
    
    while(n){
        if (n & 1) b = mul(a, b);
        a = mul(a, a);
        n >>= 1;
    }
    return b;
}

int main() {
     ll x, y, n;
    
    scanf("%lld%lld%lld", &x, &y, &n);
    mat a;
    a.a[0][0] = a.a[1][0] = 1;
    a.a[0][1] = -1;
    a.a[1][1] = 0;
   
    if (n == 1){
        printf("%lld\n", (x+mod)%mod);
    }
    else if (n == 2){
        printf("%lld\n", (y+mod)%mod);
    }
    else {
        a = pow(a, n-2);
        ll ans = ((a.a[0][0]*y+mod)%mod + (a.a[0][1]*x+mod)%mod + mod)%mod;  // 重点就是这里
        //ll ans = (a.a[0][0]*y+a.a[0][1]*x+mod)%mod;
        //ans = (ans + mod)%mod;   
        printf("%lld\n", ans );  
    }
    
    return 0;
}
/*
-9 -11
12345
*/