poj3070

用二分法求矩阵幂

先把矩阵的2^x全算出来存入数组的第x位。然后读入n，从右至左遍历数组，能乘则乘，直到把n填满。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define w 10000;

struct Pow2
{
    int matrix[2][2];
    int num;
} pow2[50];

Pow2 mul(Pow2 &a, Pow2 &b)
{
    Pow2 ret;
    for (int i = 0; i < 2; i++)
        for (int j = 0; j < 2; j++)
        {
            ret.matrix[i][j] = 0;
            for (int k = 0; k < 2; k++)
            {
                ret.matrix[i][j] += a.matrix[i][k] * b.matrix[k][j] % w;
                ret.matrix[i][j] %= w;
            }
        }
    return ret;
}

int work(int n)
{
    if (n <= 1)
        return n;
    n--;
    Pow2 temp;
    int i = 1;
    while (n >= pow2[i].num)
        i++;
    i--;
    temp = pow2[i];
    n -= pow2[i].num;
    while (n)
    {
        while (n < pow2[i].num)
            i--;
        n -= pow2[i].num;
        temp = mul(pow2[i], temp);
    }
    return temp.matrix[0][0];
}

int main()
{
    int n;
    //freopen("D:\\t.txt", "r", stdin);
    pow2[1].matrix[0][0] = 1;
    pow2[1].matrix[1][0] = 1;
    pow2[1].matrix[0][1] = 1;
    pow2[1].matrix[1][1] = 0;
    pow2[1].num = 1;
    for (int i = 1; i <= 40; i++)
        pow2[i + 1] = mul(pow2[i], pow2[i]);
    for (int i = 1; i <= 40; i++)
        pow2[i + 1].num = pow2[i].num * 2;
    while (scanf("%d", &n) != EOF && n != -1)
        printf("%d\n", work(n));
    return 0;
}