UVA 10229 Modular Fibonacci

斐波那契取MOD。利用矩阵快速幂取模

http://www.cnblogs.com/Commence/p/3976132.html

代码：

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}
struct node
{
    LL mat[2][2];
}ans,t;
LL MOD,N;
int M;
void init()
{
    ans.mat[0][0]=1;ans.mat[0][1]=0;ans.mat[1][0]=0;ans.mat[1][1]=1;
    t.mat[0][0]=1;t.mat[0][1]=1;t.mat[1][0]=1;t.mat[1][1]=0;
    MOD=1<<M;
}
node mult(node a,node b)
{
    node res;
    for (int i=0;i<2;i++)
        for (int j=0;j<2;j++)
        res.mat[i][j]=(a.mat[i][0]*b.mat[0][j]+a.mat[i][1]*b.mat[1][j])%MOD;
    return res;
}
int main()
{
    while (scanf("%lld %d",&N,&M)!=EOF)
    {
        init();
        //printf("%lld
",MOD);
        while (N)
        {
            if (N&1) ans=mult(ans,t);
            t=mult(t,t);
            N>>=1;
        }
        //printf("%lld %lld %lld %lld
",ans.mat[0][0],ans.mat[0][1],ans.mat[1][0],ans.mat[1][1]);
        printf("%lld
",(ans.mat[0][1])%MOD);
    }
    return 0;
}