2017 ECJTU ACM程序设计竞赛 矩阵快速幂+二分

矩阵

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 7   Accepted Submission(s) : 4

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Problem Description

假设你有一个矩阵，有这样的运算A^(n+1) = A^(n)*A (*代表矩阵乘法)
现在已知一个n*n矩阵A,S = A+A^2+A^3+...+A^k,输出S，因为每一个元素太大了，输出的每个元素模10

Input

先输入一个T(T<=10)，每组一个n,k(1<=n<=30, k<=1000000)

Output

输出一个矩阵,每个元素模10(行末尾没有多余空格)

Sample Input

1
3 2
0 2 0
0 0 2
0 0 0

Sample Output

0 2 4
0 0 2
0 0 0

矩阵快速幂 + 等比数列二分求和
AC

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <queue>
#include <vector>
using namespace std;
const int maxn= 1e5+10;
const double eps= 1e-6;
const int inf = 0x3f3f3f3f;
const int mod =10;
typedef long long ll;
int n,m;
struct matrix
{
    int m[35][35];
    matrix()
    {
        memset(m,0,sizeof(m));
    }
};
matrix operator *(const matrix &a,const matrix &b)
{
    matrix c;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            c.m[i][j]=0;
            for(int k=1;k<=n;k++)
                c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%mod;
        }
        return c;
}
matrix quick(matrix base,int pow)
{
    matrix a;
    for(int i=1;i<=n;i++) a.m[i][i]=1;
    while(pow)
    {
        if(pow&1) a=a*base;
        base=base*base;
        pow>>=1;
    }
    return a;
}
matrix sum(matrix a,matrix b)
{
    for(int j=1;j<=n;j++)
        for(int k=1;k<=n;k++)
            a.m[j][k]=(a.m[j][k]+b.m[j][k])%mod;
    return a;
}
matrix solve(matrix x,int y)
{
    matrix a,s;
    if(y==1)
        return x;
    a=solve(x,y/2);
    s=sum(a,a*quick(x,y/2));
    if(y&1)
        s=sum(s,quick(x,y));
    return s;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&m);
        matrix  a;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                scanf("%d",&a.m[i][j]);
        matrix ans=solve(a,m);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(j==n)
                    printf("%d
",ans.m[i][j]);
                else
                    printf("%d ",ans.m[i][j]);
            }
        }

    }
}