Ultraviolet A 11885-Number of Battlefields (matrix power)

Question connection: Ultraviolet A 11885-Number of Battlefields

The circumference p is given, and the circumference of the smallest surrounding matrix of the graph is p, excluding the rectangle.

Solution: the rapid power of the matrix. If a rectangle is contained, the number of the rectangles is equal to the even number of the Fibonacci series.

#include 
  
   #include 
   
    using namespace std;typedef long long ll;const ll MOD = 987654321;void mul(ll a[2][2], ll b[2][2], ll c[2][2]) {    ll ans[2][2];    memset(ans, 0, sizeof(ans));    for (int i = 0; i < 2; i++) {        for (int j = 0; j < 2; j++) {            for (int k = 0; k < 2; k++)                ans[i][j] = (ans[i][j] + a[i][k] * b[k][j]) % MOD;        }    }    memcpy(c, ans, sizeof(ans));}void power (ll a[2][2], int n) {    ll ans[2][2] = {1, 0, 1, 0};    while (n) {        if (n&1)            mul(ans, a, ans);        mul(a, a, a);        n /= 2;    }    memcpy(a, ans, sizeof(ans));}int main () {    int p;    while (scanf("%d", &p) == 1 && p) {        if (p&1 || p < 6) {            printf("0\n");            continue;        }        p = (p - 4) / 2;        ll a[2][2] = {1, 1, 1, 0};        /*        power(a, 2*p-1);        printf("%lld\n", (a[0][0] + a[0][1]  - p - 1 + MOD) % MOD);        */        power(a, 2*p);        printf("%lld\n", (a[1][0] - p - 1 + MOD) % MOD);    }    return 0;}