poj 3070 斐波拉切快速幂公式

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 14923 Accepted: 10496

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 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 F_n.

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

Sample Input

099999999991000000000-1

Sample Output

0346266875

涨姿势了不是，斐波拉切的快速幂公式就在这里，解决了n很大是递归公式爆栈的问题吧

注意n=0特判即可

#include <iostream>#include <cstdio>#include <cstring>using namespace std;typedef long long ll;const int maxn = 4;const int mod = 10000;struct mat {    int s[maxn][maxn];    mat(){        memset(s,0,sizeof(s));    };    mat operator * (const mat& c) {    mat ans;    for (int i = 0; i < maxn; i++) //矩阵乘法        for (int j = 0; j < maxn; j++)            for (int k = 0; k < maxn; k++)                ans.s[i][j] = (ans.s[i][j] + s[i][k] * c.s[k][j]) % mod;    return ans;    }}str;mat pow_mod(ll k) {    if (k == 1)        return str;    mat a = pow_mod(k/2);//不能改    mat ans = a * a;    if (k & 1)        ans = ans * str;    return ans;}int main() {    //freopen("in.txt","r",stdin);    int n;    str.s[0][0] = 1;    str.s[0][1] = 1;    str.s[1][0] = 1;    str.s[1][1] = 0;    while(~scanf("%d",&n)&&n!=-1) {        if(n==0)            puts("0");        else {            mat sub = pow_mod(n);            ll res = 0;            res = sub.s[0][1]%mod;            cout<<res<<endl;        }    }    return 0;}