注：机试以解题为主，优化其次

快速排序

int partition(int a[], int low, int high){
    int pivot = a[low];
    while (low <= high){
        while (a[high] > pivot && low <= high) high -- ;
        a[low] = a[high];
        while (a[low] < pivot && low <= high ) low ++ ;
        a[high] = a[low];
    }
    a[low] = pivot;
    return low;
}

二分

int l = 0, r = n - 1;
while (l < r){
    int mid = r + l >> 1;
    if (a[mid] >= x) r = mid;
    else l = mide + 1;
}

while (l < r){
    int mid = r + l + 1 >> 1;
    if (a[mid] <= x) l = mid;
    else r = mid - 1;
}

前缀和

int a[N];//保存每个元素的值
for (int i = 1; i <= n; i ++ ) s[i] = s[i - 1] + a[i];

int s[N];//元素值和前缀和用一个数组保存
for (int i = 2; i <= n; i ++ ) s[i] += s[i - 1];
//从第r个数到第l个数的区间和为 s[l] - s[r - 1]

双指针

//可利用计数数组来记录元素出现的次数
//利用快指针和慢指针

关于二进制

int lowbit(int x){
    return x & -x;  //返回x的二进制表示中最低位1出现的位置
}

void bit1(int x)  //输出某数的二进制表示
{
    int b = x;
    while (b >= 1)
    {
        p[i] = x >> i & 1;
        b = b >> 1;
        i++;
    }

    for (int j = i - 1; j >= 0; j--)
        cout << p[j];
}

进制转换

int main(){
    cin >> x;
    stack<int> s;

    if (x == 0) cout << 0;
    else {
        while (x){
            s.push(x % n);  //n表示的是转换为n进制
            x /= n;
        }
    }

    while (!s.empty()){
        cout << s.top();
        s.pop();
    }
}