【矩阵快速幂+二分】Matrix Power Series POJ

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Think:
1知识点：矩阵快速幂+二分求解等比矩阵前n项和
2题意：输入一个矩阵，求解矩阵前n项和(S = A^1 + A^2 + A^3 + … + A^k.)，模mod

vjudge题目链接

以下为Accepted代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 40;typedef struct Matrax{    int m[N][N];}matrax;matrax e, per;int n, m, mod;void read();/*读取矩阵+初始化单位矩阵*/matrax Add(matrax a, matrax b);/*矩阵加法*/matrax multi(matrax a, matrax b);/*矩阵乘法*/matrax power(int k);/*矩阵快速幂*/matrax matrax_sum(int k);/*等比矩阵前n项和*/void Pri(matrax ans);/*输出矩阵*/int main(){    while(~scanf("%d %d %d", &n, &m, &mod)){        read();        matrax ans = matrax_sum(m);        Pri(ans);    }    return 0;}void read(){/*读取矩阵+初始化单位矩阵*/    for(int i = 0; i < n; i++){        for(int j = 0; j < n; j++){            scanf("%d", &e.m[i][j]);            e.m[i][j] %= mod;            per.m[i][j] = (i == j);        }    }}matrax Add(matrax a, matrax b){/*矩阵加法*/    matrax c;    for(int i = 0; i < n; i++){        for(int j = 0; j < n; j++){            c.m[i][j] = (a.m[i][j] + b.m[i][j])%mod;        }    }    return c;}matrax multi(matrax a, matrax b){/*矩阵乘法*/    matrax c;    for(int i = 0; i < n; i++){        for(int j = 0; j < n; j++){            c.m[i][j] = 0;            for(int k = 0; k < n; k++){                c.m[i][j] += (a.m[i][k] * b.m[k][j])%mod;            }            c.m[i][j] %= mod;        }    }    return c;}matrax power(int k){/*矩阵快速幂*/    matrax p, ans = per;    p = e;    while(k){        if(k & 1)            ans = multi(ans, p);        p = multi(p, p);        k >>= 1;    }    return ans;}matrax matrax_sum(int k){/*等比矩阵前n项和*/    if(k == 1)        return e;    matrax tmp, b;    tmp = matrax_sum(k>>1);    if(k & 1){        b = power((k>>1)+1);        tmp = Add(tmp, multi(b, tmp));        tmp = Add(tmp, b);    }    else {        b = power(k>>1);        tmp = Add(tmp, multi(b, tmp));    }    return tmp;}void Pri(matrax ans){/*输出矩阵*/    for(int i = 0; i < n; i++){        for(int j = 0; j < n; j++){            printf("%d%c", ans.m[i][j], j == n-1? '\n': ' ');        }    }}

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