• hdu 4291 a short problem


    利用矩阵法求f(n),其中f(n)=3*f(n-1)+f(n-2) f(0)=0,f(1)=1

    接着就是求循环节,即可。

    求循环节:

    #include<iostream>
    #include<stdio.h>
    using namespace std;
    typedef long long LL;
    const LLMOD = 1000000007LL;
    int main() {
        LL f0 = 0, f1 = 1, temp = -1;
        for (LL i = 1;; i++) {
            temp = (3 * f1 + f0) % MOD;
            f0 = f1;
            f1 = temp;
            if (f0 == 0 && f1 == 1) {
                printf("%I64d\n", i);
                break;
            }
        }
        return 0;
    }
    
    //const LL MOD = 1000000007LL;
    //222222224
    
    //const LL MOD = 222222224LL;
    //183120
    import java.io.BufferedInputStream;
    import java.util.Scanner;
    
    public class Main {
        public static void main(String[] args) {
            Scanner cin = new Scanner(new BufferedInputStream(System.in));
            long n = -1;
            while (cin.hasNext()) {
                n = cin.nextLong();
                System.out.println(cal(cal(cal(n, 183120), 222222224), 1000000007));
            }
            cin.close();
        }
    
        static long numa[][] = new long[2][2];
        static long numb[][] = new long[2][2];
    
        public static void matrix(long[][] a, long[][] b, int mark, long MOD) {
            long[][] temp = new long[2][2];
            temp[0][0] = (a[0][0] * b[0][0] % MOD + a[0][1] * b[1][0] % MOD) % MOD;
            temp[0][1] = (a[0][0] * b[0][1] % MOD + a[0][1] * b[1][1] % MOD) % MOD;
            temp[1][0] = (a[1][0] * b[0][0] % MOD + a[1][1] * b[1][0] % MOD) % MOD;
            temp[1][1] = (a[1][0] * b[0][1] % MOD + a[1][1] * b[1][1] % MOD) % MOD;
            if (mark == 1) {
                for (int i = 0; i < 2; i++) {
                    for (int j = 0; j < 2; j++) {
                        numa[i][j] = temp[i][j];
                    }
                }
            } else {
                for (int i = 0; i < 2; i++) {
                    for (int j = 0; j < 2; j++) {
                        numb[i][j] = temp[i][j];
                    }
                }
            }
        }
    
        public static long cal(long n, long MOD) {
            numa[0][0] = 1;
            numa[0][1] = 0;
            numa[1][0] = 0;
            numa[1][1] = 1;
            numb[0][0] = 3;
            numb[0][1] = 1;
            numb[1][0] = 1;
            numb[1][1] = 0;
    
            while (n != 0) {
                if (n % 2 == 1) {
                    matrix(numa, numb, 1, MOD);
                }
                matrix(numb, numb, 0, MOD);
                n >>= 1;
            }
            return numa[0][1];
        }
    }
    /*
     * Main.cpp
     *
     *  Created on: 2012-9-16
     *      Author: wangzhu
     */
    #include<iostream>
    #include<stdio.h>
    #include<math.h>
    using namespace std;
    #define LL long long
    LL numa[2][2], numb[2][2];
    void matrix(LL a[2][2], LL b[2][2], int mark, LL MOD) {
        LL temp[2][2];
        temp[0][0] = (a[0][0] * b[0][0] % MOD + a[0][1] * b[1][0] % MOD) % MOD;
        temp[0][1] = (a[0][0] * b[0][1] % MOD + a[0][1] * b[1][1] % MOD) % MOD;
        temp[1][0] = (a[1][0] * b[0][0] % MOD + a[1][1] * b[1][0] % MOD) % MOD;
        temp[1][1] = (a[1][0] * b[0][1] % MOD + a[1][1] * b[1][1] % MOD) % MOD;
        if (mark) {
            for (int i = 0; i < 2; i++) {
                for (int j = 0; j < 2; j++) {
                    numa[i][j] = temp[i][j];
                }
            }
        } else {
            for (int i = 0; i < 2; i++) {
                for (int j = 0; j < 2; j++) {
                    numb[i][j] = temp[i][j];
                }
            }
        }
    }
    LL cal(LL n, LL MOD) {
        numa[0][0] = 1;
        numa[0][1] = 0;
        numa[1][0] = 0;
        numa[1][1] = 1;
        numb[0][0] = 3;
        numb[0][1] = 1;
        numb[1][0] = 1;
        numb[1][1] = 0;
    
        while (n) {
            if (n & 1) {
                matrix(numa, numb, 1, MOD);
            }
            matrix(numb, numb, 0, MOD);
            n >>= 1;
        }
        return numa[0][1];
    }
    int main() {
    #ifndef ONLINE_JUDGE
        freopen("data.in", "r", stdin);
    #endif
                LL n;
                while (~scanf("%I64d", &n)) {
                    printf("%I64d\n",
                            cal(cal(cal(n, 183120LL), 222222224LL),1000000007LL));
            }
            return 0;
    }
  • 相关阅读:
    Java和JavaScript的时间互传
    session.createQuery()不执行和java.lang.reflect.InvocationTargetException
    [转载]标签a的href和onclick
    [转载]前端优化指南
    POJ1328-Radar Installation
    POJ1323-Game Prediction
    codinglife主题小修改和有意思的博客挂件
    POJ1050-To the Max
    HDU4323-Magic Number(levenshtein distance-编辑距离)
    HDU2955-Robberies
  • 原文地址:https://www.cnblogs.com/xiaoxian1369/p/2687881.html
Copyright © 2020-2023  润新知