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cf 450b 矩陣快速冪(數論取模 一大坑點啊)

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

技術分享

You are given x and y, please calculate fn 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).

題意 : 顯然是矩陣快速冪麽

坑點 : 就是數論取模這一塊 , 稍不註意就錯了

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;  // 重點就是這裏
        //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
*/

cf 450b 矩陣快速冪(數論取模 一大坑點啊)