int main(void)
{
/*经过优化后的递归法求m的n次幂*/
int m;
int n;
printf("请输入mn的值：");
scanf("%d%d",&m,&n);
printf("%ld",MyPow(m,n));
getch();
return 0;
}
unsigned long MyPow(int m, int n)
{
unsigned long tmp;
if (0 == n)
{
return 1;
}
if (1 == n)
{
return m;
}
if (0==n%2)
{
tmp = MyPow(m,n/2);
return tmp*tmp;
}
if (0 != n%2)
{
return m*MyPow(m,n-1);
}
}
void fillMatrix(int n, double **t, double x0) {
int i, j;
for (j = 0; j < n; j++) {
t[0][j] = 1;
for (i = 1; i < n; i++) {
if (i + j < n)t[i][j] = t[i - 1][j] * (n - i - j);
}
}
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
if (i + j < n)t[i][j] *= MyPow(x0, n - j - i - 1);
}
}
}
DWORD CharToDword(TCHAR* start, DWORD length)
{
TCHAR temp;
DWORD i;
DWORD re = 0;
for (i = 0; i < length; i++)
{
temp = start[i];
if ('0' <= temp && '9' >= temp)
{
re += ((temp - '0')*MyPow(10, length - i - 1));
}
else
{
return 0;
}
}
return re;
}
double MyPow(double x0, int exp) {
return exp > 0 ? MyPow(x0, exp - 1) * x0 : 1;
}