POJ 3233(矩阵快速幂)

WA代码:

<pre name="code" class="cpp">

#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<queue>
#include<vector>
#include<map>
#include<iostream>
using namespace std;
int n,k,mod;
struct matrix
{
int m[32][32];
};
matrix multi(matrix a,matrix b)
{
matrix tmp;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
tmp.m[i][j]=0;
for(int z=0;z<n;z++)
tmp.m[i][j]=(tmp.m[i][j]+a.m[i][z]*b.m[z][j])%mod;
}
return tmp;
}
matrix fastmod(matrix use,int x)
{
matrix ans;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
if(i==j) ans.m[i][j]=1;
else ans.m[i][j]=0;
}
while(x)
{
if(x&1) ans=multi(ans,use);
use=multi(use,use);
x>>=1;
}
return ans;
}
int main()
{
//freopen("test.txt","r",stdin);
int t1,t2,t3;
while(scanf("%d%d%d",&t1,&t2,&t3)!=EOF)
{
matrix base,out;
memset(base.m,0,sizeof(base.m));
memset(out.m,0,sizeof(out.m));
n=t1,k=t2,mod=t3;//维数,次数,模
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
scanf("%d",&base.m[i][j]);
out.m[i][j]=base.m[i][j];
}
out=fastmod(out,k);
base=multi(base,out);
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
printf("%d",base.m[i][j]);
if(j!=n-1) printf(" ");
}
printf("\n");
}
}
}

最后参考了解答:

#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<queue>
#include<vector>
#include<map>
#include<iostream>
using namespace std;
int n,k,mod;
struct matrix
{
int m[64][64];
};
matrix multi(matrix a,matrix b)
{
matrix tmp;
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
tmp.m[i][j]=0;
for(int z=0;z<n;z++)
{
if(!a.m[i][z]||!b.m[z][j]) continue;
tmp.m[i][j]=(tmp.m[i][j]+a.m[i][z]*b.m[z][j])%mod;
}
}
}
return tmp;
}
matrix fastmod(matrix use,int x)
{
matrix ans;
memset(ans.m,0,sizeof(ans.m));
for(int i=0;i<n;i++)ans.m[i][i]=1;
while(x)
{
if(x&1) ans=multi(ans,use);
use=multi(use,use);
x>>=1;
}
return ans;
}
int main()
{
//freopen("test.txt","r",stdin);
int t1,t2,t3;
while (scanf("%d%d%d",&t1,&t2,&t3)!=EOF)
{
n=t1,k=t2,mod=t3;//维数,次数,模
matrix base,out;
memset(base.m,0,sizeof(base.m));
memset(out.m,0,sizeof(out.m));
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
scanf("%d",&base.m[i][j+n]);
out.m[i+n][j+n]=base.m[i][j+n];
}
for(int i=0;i<n;i++)
out.m[i][i]=out.m[i+n][i]=1;
n*=2;
out=fastmod(out,k);
base=multi(base,out);
for(int i=0;i<n/2;i++)
{
printf("%d",base.m[i][0]);
for(int j=1;j<n/2;j++)
printf(" %d",base.m[i][j]);
printf("\n");
}
}
return 0;
}

/*
s[k]=A+A^2+...+A^k
矩阵：
|s[0] A|*|I O|^k=|s[k] A^(k+1)|  (I为单位矩阵，O为0矩阵，A为输入的矩阵)
|O    O| |I A|   |O    O      |
*/