POJ 3070 Fibonacci 矩阵快速幂

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题意很明确，求第m个斐波那契数MOD10000的结果

题目连矩阵都构造好了，就是

1 1

1 0

然后对这个求幂就行了

/*ID: sdj22251PROG: subsetLANG: C++*/#include <iostream>#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cctype>#include <string>#include <cstring>#include <cmath>#include <ctime>#define MAXN 305#define INF 100000000#define eps 1e-7#define PI 3.1415926535898using namespace std;int n = 2, m;int tt[2][2]={{1, 1}, {1, 0}};struct wwj{    int r, c;    int mat[3][3];} need, ready;void init(){    memset(need.mat, 0, sizeof(need.mat));    need.r = n;    need.c = n;    for(int i = 1; i <= n; i++)    {        need.mat[i][i] = 1;    }    ready.c = n;    ready.r = n;    for(int i = 1; i <= n; i++)    {        for(int j = 1; j <= n; j++)        ready.mat[i][j] = tt[i - 1][j - 1];    }}wwj multi(wwj x, wwj y){    wwj t;    int i, j, k;    memset(t.mat, 0, sizeof(t.mat));    t.r = x.r;    t.c = y.c;    for(i = 1; i <= x.r; i++)    {        for(k = 1; k <= x.c; k++)            if(x.mat[i][k])            {                for(j = 1; j <= y.c; j++)                {                    t.mat[i][j] += (x.mat[i][k] * y.mat[k][j]) % 10000;                    t.mat[i][j] %= 10000;                }            }    }    return t;}int main(){    int m;    while(scanf("%d", &m) != EOF)    {        if(m == -1) break;        init();        while(m)        {            if(m & 1)            {                need = multi(ready, need);            }            ready = multi(ready, ready);            m = m >> 1;        }        printf("%d\n", need.mat[1][2] % 10000);    }    return 0;}