51Nod 1242:斐波那契数列的第N项

题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1242

N比较大，用矩阵快速幂解决。

与POJ 3070题目一样。

AC代码：

#include <iostream>#include <stdio.h>#include <math.h>#include <string.h>using namespace std;typedef long long LL;const int N = 5;LL temp[N][N];LL res[N][N];const int mod = 1000000009;void Mul(LL a[][N],LL b[][N]){    memset(temp,0,sizeof(temp));    for(int i = 0; i < 2; i++)  ///i行        for(int j = 0; j < 2; j++)  ///j列            for(int k = 0; k < 2; k++)                temp[i][j] = (temp[i][j]+(a[i][k]*b[k][j])%mod)%mod;    for(int i = 0; i < 2; i++)        for(int j = 0; j < 2; j++)            a[i][j] = temp[i][j];}void Solve(LL a[][N],LL n)  ///n是求的幂次{    memset(res,0,sizeof(res));    for(int i = 0; i < 2; i++)        res[i][i] = 1;    while(n)    {        if(n&1)            Mul(res,a);        Mul(a,a);        n>>=1;    }}int main(){    LL T;    while(~scanf("%lld",&T))    {        LL a[N][N];        a[0][0] = 1;        a[0][1] = 1;        a[1][0] = 1;        a[1][1] = 0;        if(T == 0 || T == 1) printf("%lld\n",T);        else        {            Solve(a,T);            printf("%lld\n",res[0][1]);        }    }    return 0;}