HDU - 1021ACM

这个题目很简单了，就是斐波那契数列~

至于直接加的方法应该会超时的（没试，不过应该会超时）。

所以用矩阵快速幂的方法（假如不知道快速幂怎么做，请看我上面的博客：http://blog.youkuaiyun.com/alps1992/article/details/42131581）

就比较简单了~ 其实就是构造矩阵。

代码如下：

//
//  main.cpp
//  hdu_1021
//
//  Created by Alps on 14/12/28.
//  Copyright (c) 2014年 chen. All rights reserved.
//
//  Fibonacci Again
// There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).
//Input
//Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).

//Output
//Print the word "yes" if 3 divide evenly into F(n).
//Print the word "no" if not.

#include <iostream>
using namespace std;

int MultiMatrix(int a[2][2], int b[2][2]){
    int temp[2][2];
    for (int i = 0; i < 2; i++) {
        for (int j = 0; j < 2; j++) {
            temp[i][j] = a[i][0]*b[0][j] + a[i][1]*b[1][j];
        }
    }
    for (int i = 0; i < 2; i++) {
        for (int j = 0; j < 2; j++) {
            a[i][j] = temp[i][j];
        }
    }
    return a[0][0];
}

int main(int argc, const char * argv[]) {
    int n;
    int ans;
    while (scanf("%d",&n) != EOF) {
        if (n == 0 || n == 1) {
            printf("no\n");
            continue;
        }
        n -= 1;
        int cp[2][2] = {1,0,0,1};
        int init[2][2] = {1,1,1,0};
        int matrix[2][2] = {11,7,7,4};
        while (n & 1) {
            MultiMatrix(cp, init);
            MultiMatrix(init, init);
            n = n>>1;
        }
        ans = MultiMatrix(matrix, cp);
        if (ans % 3 == 0) {
            printf("yes\n");
        }else{
            printf("no\n");
        }
    }
    return 0;
}