UVA 10229 Modular Fibonacci

UVA_10229

    将斐波那契数用矩阵形式表示，然后再做快速幂取模即可。

#include<stdio.h>
#include<string.h>
#define MAXD 50
long long int N, M, D, a[MAXD][4], b[4];
void pow_mod(long long int n, int e)
{
    if(n == 1)
    {
        a[e][0] = a[e][1] = a[e][2] = 1;
        a[e][3] = 0;
        return ;
    }
    pow_mod(n / 2, e + 1);
    a[e][0] = (a[e + 1][0] * a[e + 1][0] + a[e + 1][1] * a[e + 1][2]) % D;
    a[e][1] = (a[e + 1][0] * a[e + 1][1] + a[e + 1][1] * a[e + 1][3]) % D;
    a[e][2] = (a[e + 1][2] * a[e + 1][0] + a[e + 1][3] * a[e + 1][2]) % D;
    a[e][3] = (a[e + 1][2] * a[e + 1][1] + a[e + 1][3] * a[e + 1][3]) % D;
    if(n % 2)
    {
        b[0] = (a[e][0] * a[0][0] + a[e][1] * a[0][2]) % D;
        b[1] = (a[e][0] * a[0][1] + a[e][1] * a[0][3]) % D;
        b[2] = (a[e][2] * a[0][0] + a[e][3] * a[0][2]) % D;
        b[3] = (a[e][2] * a[0][1] + a[e][3] * a[0][3]) % D;
        a[e][0] = b[0], a[e][1] = b[1], a[e][2] = b[2], a[e][3] = b[3];
    }
}
void solve()
{
    int i;
    D = 1;
    for(i = 0; i < M; i ++)
        D <<= 1;
    pow_mod(N - 1, 1);
    printf("%lld\n", a[1][0]);
}
int main()
{
    a[0][0] = a[0][1] = a[0][2] = 1;
    a[0][3] = 0;
    while(scanf("%lld%lld", &N, &M) == 2)
    {
        if(N == 0 || M == 0)
            printf("0\n");
        else if(N == 1)
            printf("1\n");
        else
            solve();
    }
    return 0;
}