For ( i=0; i<n; i++ )

For ( j=0; j<m; j++ )

A[i][j]=0;

S=0;

For ( i=0; i<n; i++ )

For( j=0; j<n; j++ )

S+=B[i][j];

Sum=s;

I=1;

While( i<=n )

I=i∗3;

X=0;

For( i=1; i<n; i++ )

For ( j=1; j<=n-i; j++ )

X++;

下面算法的时间复杂度为 ▁▁▁▁▁。

Int foo(int n)

{

Return n ∗ (n + 1) / 2;

}

X = 0;

While ( n >= (x+1)∗(x+1) )

X = x+1;