NBUT 1481

题意
问题描述
F0=0, F1=1, F2=1, F3=2, F4=3, F5=5……
Fn = F(n-1) + F(n-2) ( n >= 2)
输入
输入一个整数n( 0 <= n <= 1,000,000,000)
输出

输出 Fn mod 20130719

#include<stdio.h>
int f[1000000];
int main()
{
    int a,b,c,i,j;
    f[0]=0;
    f[1]=1;
    for(i=2; i<=572889; i++)
        f[i]=(f[i-1]+f[i-2])%20130719;
    while(scanf("%d",&a)!=EOF)
    {
        printf("%d\n",f[a%572880]);
    }
    return 0;
}
矩阵乘法快速幂：
#include<stdio.h>
#define N 20130719
struct matrix
{
    long long s[3][3];
} A, B, D;

matrix mul(matrix D, matrix A)
{
    int i, j, k;
    matrix s;
    for(i = 0; i < 2; i++)
        for(j = 0; j < 2; j++)
        {
            s.s[i][j] = 0;
            for(k = 0; k < 2; k++)
                s.s[i][j] = (s.s[i][j] + D.s[i][k] * A.s[k][j]) % N;
        }
    return s;
}

matrix pow(matrix A, long long k)
{
    matrix s , D = A;
    s.s[0][0] = 1;
    s.s[0][1] = 0;
    s.s[1][0] = 0;
    s.s[1][1] = 1;
    while(k > 0)
    {
        if(k & 1) s = mul(D, s);
        D = mul(D, D);
        k = k >> 1;
    }
    return s;
}
int main ()
{
    long long n;
    while(scanf("%lld", &n) != EOF)
    {
        if(n == 0)
        {
            printf("0\n");
            continue;
        }
        if(n == 1)
        {
            printf("1\n");
            continue;
        }
        A.s[0][0] = 1;
        A.s[0][1] = 1;
        A.s[1][0] = 1;
        A.s[1][1] = 0;
        B = pow(A,n);
        printf("%d\n",B.s[0][1]%N);
    }
    return 0;
}