POJ 3070 Fibonacci

题目描述：

In the Fibonacci integer sequence, , , and  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 .

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:

题目大意

求斐波那契第n项。

解题报告：

1：矩阵已列出，直接套模板就行了。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const ll N = 2;
const ll mod = 10000;
struct matrix{
    ll num[N][N];
}A, B;
void init(){
    memset(A.num, 0, sizeof(A.num));
    memset(B.num, 0, sizeof(B.num));
    A.num[0][0] = A.num[0][1] = A.num[1][0] = 1;
}
matrix mul(matrix AA, matrix BB){
    matrix C;
    memset(C.num, 0, sizeof(C.num));
    for(ll i=0; i<N; ++i){
        for(ll j=0; j<N; ++j){
            for(ll k=0; k<N; ++k){
                C.num[i][j] += AA.num[i][k]*BB.num[k][j];
                C.num[i][j] %= mod;
            }
        }
    }
    return C;
}
ll Pow(ll n){
    B.num[0][0] = 1, B.num[1][0] = 0;
    while(n){
        if(n&1)B = mul(A, B);
        n >>= 1, A = mul(A, A);
    }
    return B.num[0][0]%mod;
}
int main(){
    ll n;
    while(~scanf("%lld", &n) && n != -1){
        init();
        if(n == 0){
            printf("0\n");
            continue;
        }
        printf("%lld\n", Pow(n-1));
    }
    return 0;
}

 

 

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标签：matrix,ll,3070,num,POJ,Fibonacci,test,include
来源： https://blog.csdn.net/jun_____/article/details/104072594