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  • POJ3070 Fibonacci

    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 13731   Accepted: 9727

    Description

    In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

    0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

    An alternative formula for the Fibonacci sequence is

    .

    Given an integer n, your goal is to compute the last 4 digits of Fn.

    Input

    The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.

    Output

    For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).

    Sample Input

    0
    9
    999999999
    1000000000
    -1

    Sample Output

    0
    34
    626
    6875

    Hint

    As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

    .

    Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

    .

    Source

    矩阵乘法入门题。

    hint已经写明了解法

     1 /*by SilverN*/
     2 #include<algorithm>
     3 #include<iostream>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<cmath>
     7 #include<vector>
     8 using namespace std;
     9 const int mod=1e4;
    10 const int mxn=240;
    11 struct mat{
    12     int s[3][3];
    13     mat(){memset(s,0,sizeof s);}
    14 };
    15 mat operator * (const mat a,const mat b){
    16     mat c;
    17     for(int i=1;i<=2;i++)
    18      for(int j=1;j<=2;j++)
    19        for(int k=1;k<=2;k++){
    20         (c.s[i][j]+=a.s[i][k]*b.s[k][j])%=mod;
    21        }
    22     return c;
    23 }
    24 int n;
    25 int main(){
    26     int i,j;
    27     while(scanf("%d",&n) && n!=-1){
    28         mat a,b;
    29         a.s[1][1]=1;a.s[1][2]=1;a.s[2][1]=1;a.s[2][2]=0;
    30         b.s[1][1]=1;b.s[1][2]=0;b.s[2][1]=0;b.s[2][2]=1;
    31         for(;n;n>>=1,a=a*a)
    32             if(n&1)b=b*a;
    33         printf("%d
    ",b.s[2][1]);
    34     }
    35     return 0;
    36 }
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  • 原文地址:https://www.cnblogs.com/SilverNebula/p/6019661.html
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