Posted on 2010-08-23 21:23
acronix 阅读(12608)
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hzshuai收集的模板
下面是 m^n % k 的快速幂：
// m^n % k
int quickpow(int m,int n,int k)
{
int b = 1;
while (n > 0)
{
if (n & 1)
b = (b*m)%k;
n = n >> 1 ;
m = (m*m)%k;
}
return b;
}
下面是矩阵快速幂：
//HOJ 3493
/*===================================*/
|| 快速幂（quickpow）模板
|| P 为等比，I 为单位矩阵
|| MAX 要初始化！！！！
||
/*===================================*/
/*****************************************************/
#include <cstdio>
const int MAX = 3;
typedef struct{
int m[MAX][MAX];
} Matrix;
Matrix P = {5,-7,4,
1,0,0,
0,1,0,
};
Matrix I = {1,0,0,
0,1,0,
0,0,1,
};
Matrix matrixmul(Matrix a,Matrix b) //矩阵乘法
{
int i,j,k;
Matrix c;
for (i = 0 ; i < MAX; i++)
for (j = 0; j < MAX;j++)
{
c.m[i][j] = 0;
for (k = 0; k < MAX; k++)
c.m[i][j] += (a.m[i][k] * b.m[k][j])%9997;
c.m[i][j] %= 9997;
}
return c;
}
Matrix quickpow(long long n)
{
Matrix m = P, b = I;
while (n >= 1)
{
if (n & 1)
b = matrixmul(b,m);
n = n >> 1;
m = matrixmul(m,m);
}
return b;
}
/*************************************/
int main()
{
Matrix re;
int f[3] = {2,6,19};
long long n;
while (scanf("%I64d",&n) && n != 0)
{
if (n == 1)
printf("1\n");
else if (n <= 4)
printf("%d\n",f[n-2]);
else {
re = quickpow(n - 4);
printf("%d\n",(((re.m[0][0]*f[2])
+ (re.m[0][1]*f[1]) + (re.m[0][2]*f[0])) %9997 + 9997) % 9997);
}
}
return 0;
}