Number of Battlefields UVA - 11885 矩阵快速幂

题目链接:点我

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题意:

给周长，求能围成的战场数目，不包括矩形。

思路:

矩阵快速幂;
如果包含矩形的话，对应的则是斐波那契数列的偶数项，所以对应减去矩形的个数即可

代码:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long LL;
const int  mod = 987654321;

struct mat{
    LL a[3][3];
    mat(){memset(a, 0, sizeof(a)); }
    mat operator *( const mat q){
        mat c;
        for(int i = 1; i <= 2; ++i)
            for(int k = 1; k <= 2; ++k)
            if(a[i][k])
        for(int j = 1; j <= 2; ++j){
            c.a[i][j] += a[i][k] * q.a[k][j];
            if(c.a[i][j] >= mod)
                c.a[i][j] %= mod;
        }return c;
    }
};

mat qpow(mat x, LL n){
    mat ans;
    ans.a[1][1] = ans.a[2][2] = 1;
    while(n){
        if(n&1) ans = ans * x;
        x = x * x;
        n >>= 1;
    }return ans;
}

int main(){
    LL p;
    while(scanf("%lld", &p) != EOF&& p){
        if(p < 8 || p&1){
            printf("0\n");
            continue;
        }mat ans;
        ans.a[1][1] = ans.a[2][1] = ans.a[1][2] = 1;
        ans = qpow(ans, p-4);
        LL sum = (ans.a[1][1] - p/2 + 1 + mod) % mod;
        printf("%lld\n",sum);
    }return 0;
}