等比矩阵求和-POJ3233

https://vjudge.net/contest/182427#problem/B

Description

Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.

Input

The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 10^9) and m (m < 10^4). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.

Output

Output the elements of S modulo m in the same way as A is given.

Sample Input

2 2 4
0 1
1 1
Sample Output

1 2
2 3
Source

POJ Monthly–2007.06.03, Huang, Jinsong

如果k为偶数，那么(A+A^2+….A^K) = (A+…+A^K/2)+A^K/2*(A+…+A^K/2)
如果k为奇数，那么(A+A^2+….A^K) = (A+…+A^K/2)+A^K/2*(A+…+A^K/2)+A^k

S(k)=A+A^2+……+A^k
偶数: S(k)=S(k/2)+S(k/2)*A^(k/2)
奇数: S(k)=S(k/2)+S(k/2)*A^(k/2)+A^k;

直接用快速幂计算需要O（n）的时间，10^9的数据会超时
通过上面的公式解决超时问题

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>using namespace std;int n,k,mod;struct Matrix{    int a[35][35];};Matrix m,unit;Matrix add(Matrix x,Matrix y)//矩阵加法{    Matrix ans;    for(int i=0; i<n; i++)        for(int j=0; j<n; j++)        {            ans.a[i][j]=0;            ans.a[i][j]=(x.a[i][j]+y.a[i][j])%mod;        }    return ans;}Matrix multi(Matrix x,Matrix y)//矩阵乘法{    Matrix ans;    for(int i=0; i<n; i++)        for(int j=0; j<n; j++)        {            ans.a[i][j]=0;            for(int k=0; k<n; k++)                ans.a[i][j]=(ans.a[i][j]+x.a[i][k]*y.a[k][j])%mod;        }    return ans;}Matrix mqmod(int l)//快速幂{    Matrix p,q;    p=m,q=unit;    while(l)    {        if(l&1)q=multi(q,p);        p=multi(p,p);        l>>=1;    }    return q;}Matrix matrixsum(int k)//利用公式递归求解{    if(k==1)return m;    Matrix temp,tnow;    temp=matrixsum(k/2);    tnow=mqmod(k/2);    temp=add(temp,multi(temp,tnow));    if(k&1)//奇数的时候多加上A^k        temp=add(mqmod(k),temp);    return temp;}int main(){    scanf("%d%d%d",&n,&k,&mod);    int q;    for(int i=0; i<n; i++)        for(int j=0; j<n; j++)        {            scanf("%d",&q);            m.a[i][j]=q%mod;            unit.a[i][j]=(i==j);//如果i==j那么矩阵中此值就是1,否则为0,就是主对角线是1的单位矩阵        }    Matrix ans;    ans=matrixsum(k);    for(int i=0; i<n; i++)    {        for(int j=0; j<n-1; j++)            printf("%d ",ans.a[i][j]);        printf("%d\n",ans.a[i][n-1]);    }    return 0;}