BZOJ2476: 战场的数目(矩阵快速幂)

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10, mod = 987654321;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N;
int add(int x, int y) {
    if(x + y < 0) return x + y + mod;
    return x + y >= mod ? x + y - mod : x + y;
}
int mul(int x, int y) {
    return 1ll * x * y % mod;
}
struct Ma {
    int m[3][3];
    Ma() {
        memset(m, 0, sizeof(m));
    }
    Ma operator * (const Ma &rhs) const {
        Ma ans;
        for(int k = 1; k <= 2; k++)
            for(int i = 1; i <= 2; i++)
                for(int j = 1; j <= 2; j++)
                    ans.m[i][j] = add(ans.m[i][j], mul(m[i][k], rhs.m[k][j]));
        return ans;
    }
};
Ma MatrixPow(Ma a, int p) {
    Ma base;
    for(int i = 1; i <= 2; i++) base.m[i][i] = 1;
    if(p < 0) return base;
    while(p) {
        if(p & 1) base = base * a;
        a = a * a; p >>= 1;
    }
    return base;
}
int main() {
    while(scanf("%d", &N) && (N)) {
        if((N & 1) || (N < 8)) {puts("0"); continue;}
        if(N == 8) {printf("2\n"); continue;}
        N >>= 1;
        Ma a;
        a.m[1][1] = 3; a.m[1][2] =  - 1;
        a.m[2][1] = 1; a.m[2][2] = 0;
        a = MatrixPow(a, N - 5);
        printf("%d\n", add(add(mul(13, a.m[1][1]), mul(5, a.m[1][2])), -(N - 1)));
    }
    return 0;
}
/*
7
8
9
10
0
*/