hdu2157 How many ways?? (矩阵快速幂)
2013-08-28 19:01
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2157
题解:见:http://www.matrix67.com/blog/archives/276
#include <stdio.h>
#include <string.h>
#define MAXN 21
#define MOD 1000
typedef struct Matrix
{
int matr[MAXN][MAXN];
}Matrix;
int n,temp[MAXN][MAXN];
Matrix init,unit,c;
Matrix MatrixMult(Matrix a,Matrix b)
{
int i,j,k;
for (i=0;i<n;++i)
{
for(j=0;j<n;++j)
{
c.matr[i][j]=0;
for(k=0;k<n;++k)
c.matr[i][j]+=(a.matr[i][k]*b.matr[k][j])%MOD;
c.matr[i][j]%=MOD;
}
}
return c;
}
int pow_mod(int a,int b,int k)
{
int i,j;
for(i=0;i<n;++i)
{
for(j=0;j<n;++j)
{
init.matr[i][j]=temp[i][j];
unit.matr[i][j]=0;
}
unit.matr[i][i]=1;//单位矩阵
}
while(k)
{
if(k&1)
unit=MatrixMult(unit,init);
init=MatrixMult(init,init);
k>>=1;
}
return unit.matr[a][b];
}
int main()
{
int s,t,ans,i,j,k,m;
while(scanf("%d %d",&n,&m))
{
if(m+n==0)
break;
memset(temp,0,sizeof(temp));
for(i=0;i<m;++i)
{
scanf("%d %d",&s,&t);
temp[s][t]=1;
}
scanf("%d",&t);
while(t--)
{
scanf("%d %d %d",&i,&j,&k);
if(k==0&&i!=j)
printf("0\n");
else if(k==0&&i==j)
printf("1\n");
else
{
ans=pow_mod(i,j,k);
printf("%d\n",ans);
}
}
}
return 0;
}
题解:见:http://www.matrix67.com/blog/archives/276
#include <stdio.h>
#include <string.h>
#define MAXN 21
#define MOD 1000
typedef struct Matrix
{
int matr[MAXN][MAXN];
}Matrix;
int n,temp[MAXN][MAXN];
Matrix init,unit,c;
Matrix MatrixMult(Matrix a,Matrix b)
{
int i,j,k;
for (i=0;i<n;++i)
{
for(j=0;j<n;++j)
{
c.matr[i][j]=0;
for(k=0;k<n;++k)
c.matr[i][j]+=(a.matr[i][k]*b.matr[k][j])%MOD;
c.matr[i][j]%=MOD;
}
}
return c;
}
int pow_mod(int a,int b,int k)
{
int i,j;
for(i=0;i<n;++i)
{
for(j=0;j<n;++j)
{
init.matr[i][j]=temp[i][j];
unit.matr[i][j]=0;
}
unit.matr[i][i]=1;//单位矩阵
}
while(k)
{
if(k&1)
unit=MatrixMult(unit,init);
init=MatrixMult(init,init);
k>>=1;
}
return unit.matr[a][b];
}
int main()
{
int s,t,ans,i,j,k,m;
while(scanf("%d %d",&n,&m))
{
if(m+n==0)
break;
memset(temp,0,sizeof(temp));
for(i=0;i<m;++i)
{
scanf("%d %d",&s,&t);
temp[s][t]=1;
}
scanf("%d",&t);
while(t--)
{
scanf("%d %d %d",&i,&j,&k);
if(k==0&&i!=j)
printf("0\n");
else if(k==0&&i==j)
printf("1\n");
else
{
ans=pow_mod(i,j,k);
printf("%d\n",ans);
}
}
}
return 0;
}
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