Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.
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 the elements of S modulo m in the same way as A is given.
2 2 4 0 1 1 1Sample Output
1 22 3
思路:因为A+A^2+A^3......A^k可化为F(n)=A+A*F(n-1),所以可以变为
{F(k)}={F(n-1)}*{A,A}={A}*{A,A}^k-1
{ 1 } { 1 } {0,1} {1} {0,1}
或者用二分A+A^2+A^3......A^k=(A+A^2+A^3......A^k/2)*k/2*(A+A^2+A^3......A^k/2);
这题我用的是矩阵套矩阵的方法,但是不知道wa在哪里,怎么测都是对的。估计是我太菜;
祥神32msAC代码:
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
#include <math.h>
#define LL long long
#define FOR(i) for(int i=1;i<=n;i++)
using namespace std;
int d[100][100];
int m;
struct matrix
{
int n;
int a[100][100];
matrix(int nn,int flag):n(nn)
{
if(flag==0) FOR(i)FOR(j)a[i][j]=0;
if(flag==1) FOR(i)FOR(j)a[i][j]=i==j;
if(flag==2) FOR(i)FOR(j)a[i][j]=d[i][j];
}
matrix operator *(const matrix &b)const
{
matrix c(n,0);
FOR(i)FOR(j)if(a[i][j])
FOR(k)c.a[i][k]+=a[i][j]*b.a[j][k];
FOR(i)FOR(j)if(c.a[i][j]>m)c.a[i][j]%=m;
return c;
}
matrix powmat(int x)
{
matrix ans(n,0),tmp(n,2);
for(int i=n/2+1;i<=n;i++)
ans.a[i-n/2][i]=1;
for(; x; tmp=tmp*tmp,x>>=1)
if(x&1)ans=ans*tmp;
return ans;
}
};
int n;
int main()
{
int k;
while(~scanf("%d%d%d",&n,&k,&m))
{
int g;
memset(d,0,sizeof(d));
for(int i=1; i<=n; i++)
{
d[i][i]=1;
for(int j=1; j<=n; j++)
{
scanf("%d",&g);
if(g>m)g%=m;
d[i+n][j]=d[i+n][j+n]=g;
}
}
matrix w=matrix(2*n,2).powmat(k);
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
printf("%d ",w.a[i][j]%m);
printf("\n");
}
}
}
自己的找不到错误的wa代码(过些时间在回来重新做一遍):#include<algorithm>
#include<string.h>
#include<stdio.h>
#define LL long long
using namespace std;
int n,m;
struct node
{
LL f[35][35];
};
void init(node &a,node &b)
{
for(int i=1; i<=n; i++)
{
b.f[i][i]=1;
}
memset(a.f,0,sizeof(a.f));
}
void jzcj(node &c,node &a,node &b)
{
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
for(int k=1; k<=n; k++)
{
c.f[i][j]+=a.f[i][k]*b.f[k][j];
c.f[i][j]%=m;
}
}
}
}
void see(node a)
{
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
printf("%d ",a.f[i][j]);
}
printf("\n");
}
}
node jjzm(node a,int k)
{
node p[3][3],z[3][3],v[3][3],ans;
p[1][1]=p[1][2]=a;
init(p[2][1],p[2][2]);
z[1][1]=z[2][2]=p[2][2];
z[1][2]=z[2][1]=p[2][1];
while(k)
{
if(k&1)
{
for(int i=1; i<=2; i++)
{
for(int j=1; j<=2; j++)
{
memset(v[i][j].f,0,sizeof(v[i][j].f));
for(int k=1; k<=2; k++)
{
jzcj(v[i][j],p[i][k],z[k][j]);
}
}
}
for(int i=1; i<=2; i++)
{
for(int j=1; j<=2; j++)
{
z[i][j]=v[i][j];
}
}
}
k/=2;
for(int i=1; i<=2; i++)
{
for(int j=1; j<=2; j++)
{
memset(v[i][j].f,0,sizeof(v[i][j].f));
for(int k=1; k<=2; k++)
{
jzcj(v[i][j],p[i][k],p[k][j]);
}
}
}
for(int i=1; i<=2; i++)
{
for(int j=1; j<=2; j++)
{
p[i][j]=v[i][j];
}
}
}
memset(ans.f,0,sizeof(ans.f));
// see(z[1][1]);
jzcj(ans,a,z[1][1]);
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
ans.f[i][j]+=z[1][2].f[i][j];
ans.f[i][j]%=m;
}
}
return ans;
}
int main()
{
int k;
scanf("%d%d%d",&n,&k,&m);
{
node a,c;
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
scanf("%I64d",&a.f[i][j]);
}
}
c=jjzm(a,k-1);
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
printf("%I64d ",c.f[i][j]%m);
}
printf("\n");
}
}
}