poj3070 Fibonacci(矩阵快速幂)

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矩阵的快速幂算法
/*#include<iostream>using namespace std;typedef unsigned long long ll;#define MOD 10000struct Mat{};ll f[2];ll fib(int n){   //int f[2];    f[1]=0;f[2]=1;    for(int  i=2;i<=n;i++)    {        f[0]=f[1]%MOD;        f[1]=f[2]%MOD;        f[2]=(f[1]+f[0])%MOD;    }    return f[2];}int main(){   int n;    while(cin>>n&&n!=-1)    {   if(n==0)        cout<<0<<endl;        else if(n==1)            cout<<1<<endl;        else        cout<<fib(n)%MOD<<endl;    }    return 0;}*/#include <cstdio>#include <iostream>#include<cstdlib>using namespace std;typedef unsigned long long ll;const int MOD = 10000;/*struct matrix{    int m[2][2];}ans, base;matrix multi(matrix a, matrix b){    matrix tmp;    for(int i = 0; i < 2; ++i)    {        for(int j = 0; j < 2; ++j)        {            tmp.m[i][j] = 0;            for(int k = 0; k < 2; ++k)                tmp.m[i][j] = (tmp.m[i][j]+ a.m[i][k] * b.m[k][j])%MOD ;        }    }    return tmp;}int fast_mod(int n)  // 求矩阵 base 的  n 次幂{    base.m[0][0] = base.m[0][1] = base.m[1][0] = 1;    base.m[1][1] = 0;    ans.m[0][0] = ans.m[1][1] = 1;  // ans 初始化为单位矩阵    ans.m[0][1] = ans.m[1][0] = 0;    while(n)    {        if(n & 1)  //实现 ans *= t; 其中要先把 ans赋值给 tmp，然后用 ans = tmp * t        {            ans = multi(ans, base);        }            base = multi(base, base);        n >>= 1;    }    return ans.m[0][1];}*/struct Mat {    ll mat[2][2];};Mat operator * (Mat a, Mat b) {    Mat c;    memset(c.mat, 0, sizeof(c.mat));    int i, j, k;    for(k = 0; k < 2; ++k) {        for(i = 0; i < 2; ++i) {            if(a.mat[i][k] <= 0)  continue;            for(j = 0; j < 2; ++j) {                if(b.mat[k][j] <= 0)    continue;                c.mat[i][j] =( c.mat[i][j]+a.mat[i][k] * b.mat[k][j])%MOD;            }        }    }    return c;}Mat operator ^ (Mat a, int k) {    Mat c;    int i,j;    for(i=0;i<2;i++)        for(j=0;j<2;j++)        c.mat[i][j]=(i==j);    for(;k;k>>=1)    {   if(k&1) c=c*a;        a=a*a;    }    return c;}int main(){   Mat A,B;    A.mat[0][0]=1;    A.mat[0][1]=1;    A.mat[1][0]=1;    A.mat[1][1]=0;    int n;    while(cin>>n && n != -1)    {       B=A^n;        cout<<B.mat[1][0]%MOD<<endl;    }    return 0;}
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