Time Limit: 3000MS | Memory Limit: 131072K | |
Total Submissions: 22895 | Accepted: 9544 |
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 ≤ 109) and m (m < 104). 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
这个题通过观察很容易看出一个式子 S[k]=A+A*S[k-1]
由此可以推出
S[k-1] 1 * A 0 = S[k] 1 (1表示单位矩阵 任何矩阵乘单位矩阵为本身)
A 1
虽然 A是 矩阵但是没关系
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<math.h>
#include<vector>
#include<iostream>
using namespace std;
typedef long long ll;
ll n,mod,k;
struct node1 //矩阵里面的矩阵
{
ll a[31][31];
node1()
{
memset(a,0,sizeof(a));
}
};
node1 operator *(node1 a,node1 b)
{
node1 ans;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
for(int k=0;k<n;k++)
ans.a[i][j]=ans.a[i][j]+a.a[i][k]*b.a[k][j];
ans.a[i][j]=ans.a[i][j]%mod;
}
return ans;
}
node1 operator +(node1 a,node1 b)
{
node1 ans;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
ans.a[i][j]=a.a[i][j]+b.a[i][j];
return ans;
}
struct node2
{
node1 b[2][2];
}A,B;
node2 operator *(node2 a,node2 b)
{
node2 ans;
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
for(int k=0;k<2;k++)
ans.b[i][j]=ans.b[i][j]+a.b[i][k]*b.b[k][j];
return ans;
}
void aa(ll x)
{
node2 ans=A,p=B;
while(x)
{
if(x&1)
ans=ans*p;
p=p*p;
x=x/2;
}
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
printf("%lld",(ans.b[0][0].a[i][j]%mod+mod)%mod);
if(j!=n-1)
printf(" ");
}
printf("\n");
}
}
int main()
{
scanf("%lld%lld%lld",&n,&k,&mod);
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
scanf("%lld",&A.b[0][0].a[i][j]);
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
if(i==j)
A.b[0][1].a[i][j]=1;
else
A.b[0][1].a[i][j]=0;
}
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
B.b[0][0].a[i][j]=A.b[0][0].a[i][j];
B.b[1][0].a[i][j]=A.b[0][0].a[i][j];
B.b[0][1].a[i][j]=0;
if(i==j)
B.b[1][1].a[i][j]=1;
else
B.b[1][1].a[i][j]=0;
}
aa(k-1);
}