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int a[M][M], b[M][M], c[M][M];
for (int i = 0; i < ax; i++)
for (int j = 0; j < by; j++)
{
for (int k = 0; k < ay; k++)
c[i][j] += a[i][k] * b[k][j];
}我們常常會需要計算\(a^n\),但O(n)的作法太耗時間了,這時候我們就可以把n拆成二進位,並把是其中一的位元相乘即可
int power(int a, int n)
{
int ans = 1;
for (;n; n = n >> 1)
{
a = a*a;
if (n & 1) // 代表此位元是1
ans = ans*a;
}
return ans;
}