// 數(shù)組求和
int sum(int*a, int n)
{
     return n == 0 ? 0 : sum(a, n -1) + a[n -1];
}

// 求數(shù)組的最大值和最小值，返回值在maxValue和minValue
void MaxandMin(int *a, int l, int r, int& maxValue, int& minValue)
{
    if(l == r) // l與r之間只有一個(gè)元素
    {
        maxValue = a[l] ;
        minValue = a[l] ;
        return ;
    }

    if(l + 1 == r) // l與r之間只有兩個(gè)元素
    {
        if(a[l] >= a[r])
        {
            maxValue = a[l] ;
            minValue = a[r] ;
        }
        else
        {
            maxValue = a[r] ;
            minValue = a[l] ;
        }
        return ;
    }

    int m = (l + r) / 2 ; // 求中點(diǎn)

    int lmax ; // 左半部份最大值
    int lmin ; // 左半部份最小值
    MaxandMin(a, l, m, lmax, lmin) ; // 遞歸計(jì)算左半部份

    int rmax ; // 右半部份最大值
    int rmin ; // 右半部份最小值
    MaxandMin(a, m + 1, r, rmax, rmin) ; // 遞歸計(jì)算右半部份

    maxValue = max(lmax, rmax) ; // 總的最大值
    minValue = min(lmin, rmin) ; // 總的最小值
}

// 找出數(shù)組的最大值和次大值，a是待查找的數(shù)組，left和right是查找區(qū)間，max和second存放結(jié)果
void MaxandMin(int a[], int left, int right, int&max, int&second)
{
    if(left == right)
    {
        max = a[left] ;
        second =  INT_MIN;
    }
    elseif(left +1== right)
    {
        max = a[left] > a[right] ? a[left] : a[right] ;
        second = a[left] < a[right] ? a[left] : a[right] ;
    }
    else
    {
        int mid = left + (right - left) /2 ;

        int leftmax ;
        int leftsecond ;
        MaxandMin(a, left, mid, leftmax, leftsecond) ;

        int rightmax ;
        int rightsecond ;
        MaxandMin(a, mid +1, right, rightmax, rightsecond) ;

        if (leftmax > rightmax)
        {
            max = leftmax ;
            second = leftsecond > rightmax ? leftsecond : rightmax ;
        }
        else
        {
            max = rightmax ;
            second = leftmax < rightsecond ? rightsecond : leftmax ;
        }
    }
}

// 找出數(shù)組中出現(xiàn)次數(shù)超過(guò)一半的元素
int Find(int* a, int n)
{
    int curValue = a[0] ;
    int count = 1 ;

    for (int i = 1; i < n; ++i)
    {
        if (a[i] == curValue)
            count++ ;
        else
        {
            count-- ;
            if (count < 0)
            {
                curValue = a[i] ;
                count = 1 ;
            }
        }
    }

    return curValue ;
}

int compare(const void* a, const void* b)
{
    return *(int*)a - *(int*)b ;
}

// 求數(shù)組中元素的最短距離
void MinimumDistance(int* a, int n)
{
    // Sort
    qsort(a, n, sizeof(int), compare) ;

    int i ; // Index of number 1
    int j ; // Index of number 2

    int minDistance = numeric_limits<int>::max() ;
    for (int k = 0; k < n - 1; ++k)
    {
        if (a[k + 1] - a[k] < minDistance)
        {
            minDistance = a[k + 1] - a[k] ;
            i = a[k] ;
            j = a[k + 1] ;
        }
    }

    cout << "Minimum distance is: " << minDistance << endl ;
    cout << "i = " << i << " j = " << j << endl ;
}

// 找出兩個(gè)數(shù)組的共同元素
void FindCommon(int* a, int* b, int n)
{
    int i = 0;
    int j = 0 ;

    while (i < n && j < n)
    {
        if (a[i] < b[j])
            ++i ;
        else if(a[i] == b[j])
        {
            cout << a[i] << endl ;
            ++i ;
            ++j ;
        }
        else// a[i] > b[j]
            ++j ;
    }
}

// 三個(gè)數(shù)組的共同元素-只找最小的
void FindCommonElements(int a[], int b[], int c[], int x, int y, int z)
{
    for(int i = 0, j = 0, k = 0; i < x && j < y && k < z;)
    {
        if(a[i] < b[j])
        {
            i++ ;
        }
        else // a[i] >= b[j]
        {
            if(b[j] < c[k])
            {
                j++ ;
            }
            else // b[j] >= c[k]
            {
                if(c[k] < a[i])
                {
                    k++ ;
                }
                else // c[k] >= a[i]
                {
                    cout << c[k] << endl ;
                    return ;
                }
            }
        }
    }

    cout << "Not found!" << endl ;
}

// 找出三個(gè)數(shù)組的共同元素
// O(NlogN)
int UniqueCommonItem(int *a, int *b, int *c, int n)
{
    // sort array a
    qsort(a, n, sizeof(int), compare) ; // NlogN

    // sort array b
    qsort(b, n, sizeof(int), compare) ; // NlogN

    // for each element in array c, do a binary search in a and b
    // This is up to a complexity of N*2*logN
    for (int i = 0; i < n; i++)
    {
        if(BinarySearch(a, n, c[i]) && BinarySearch(b, n, c[i]))
            return c[i] ;
    }

    return - 1 ; // not found
}

// 找出三個(gè)數(shù)組唯一的共同元素
// O(NlogN)
int UniqueCommonItem1(int *a, int *b, int *c, int n)
{
    // sort array a
    qsort(a, n, sizeof(int), compare) ; // NlogN

    // Space for time
    bool *bb = new bool[n] ;
    memset(bb, 0, n) ;

    bool *bc = new bool[n] ;
    memset(bb, 0, n) ;

    // for each element in b, do a BS in a and mark all the common element
    for (int i = 0; i < n; i++) // NlogN
    {
        if(BinarySearch(a, n, b[i]))
            bb[i] = true ;
    }

    // for each element in c, do a BS only if b[i] is true
    for (int i = 0; i < n; i++) // NlogN
    {
        if(b[i] && BinarySearch(a, n, c[i]))
            return c[i] ;
    }

    return - 1 ; // not found
}

// Determine whether a contains value k
bool BinarySearch(int *a, int n, int k)
{
    int left = 0 ;
    int right = n - 1 ;
    while (left <= right)
    {
        int mid = (left + right) ;

        if(a[mid] < k)
            left = mid + 1 ;
        if(a[mid] == k)
            return true ;
        else
            right = mid - 1 ;
    }

    return false ;
}

// Compare function for qsort
int compare(const void* a, const void* b)
{
    return *(int*)a - *(int*)b ;
}

int FindElementWithOddCount(int*a, int n)
{
     int r = a[0] ;

     for (int i =1; i < n; ++i)
     {
          r ^= a[i] ;
     }

     return r ;
}

// 找出滿足給定和的數(shù)對(duì)
void FixedSum(int* a, int* b, int n, int d)
{
    for (int i = 0, j = n - 1; i < n && j >= 0)
    {
        if (a[i] + b[j] < d)
            ++i ;
        else if (a[i] + b[j] == d)
        {
            cout << a[i] << ", " << b[j] << endl ;
            ++i ;
            --j ;
        }
        else // a[i] + b[j] > d
            --j ;
    }
}

// 子數(shù)組的最大和
int Sum(int* a, int n)
{
    int curSum = 0;
    int maxSum = 0;
    for (int i = 0; i < n; i++)
    {
        if (curSum + a[i] < 0)
            curSum = 0;
        else
        {
            curSum += a[i] ;
            maxSum = max(maxSum, curSum);
        }
    }
    return maxSum;
}

// 子數(shù)組的最大乘積
int MaxProduct(int *a, int n)
{
    int maxProduct = 1; // max positive product at current position
    int minProduct = 1; // min negative product at current position
    int r = 1; // result, max multiplication totally

    for (int i = 0; i < n; i++)
    {
        if (a[i] > 0)
        {
            maxProduct *= a[i];
            minProduct = min(minProduct * a[i], 1);
        }
        else if (a[i] == 0)
        {
            maxProduct = 1;
            minProduct = 1;
        }
        else // a[i] < 0
        {
            int temp = maxProduct;
            maxProduct = max(minProduct * a[i], 1);
            minProduct = temp * a[i];
        }

        r = max(r, maxProduct);
    }

    return r;
}

// 將buffer中start和end之間的元素逆序
void Reverse( int buffer[], int start, int end )
{
    while ( start < end )
    {
        int temp = buffer[ start ] ;
        buffer[ start++ ] = buffer[ end ] ;
        buffer[ end-- ] = temp ;
    }
}

// 將含有n個(gè)元素的數(shù)組buffer右移k位
void Shift( int buffer[], int n, int k )
{
    k %= n ;

    Reverse( buffer, 0, n - k - 1) ;
    Reverse( buffer, n - k, n - 1 ) ;
    Reverse( buffer, 0, n - 1 ) ;
}

// 字符串逆序
void Reverse(char*a, int n)
{
     int left =0;
     int right = n -1;

     while (left < right)
     {
         char temp = a[left] ;
         a[left++] = a[right] ;
         a[right--] = temp ;
     }
}

// n選m的所有組合
int buffer[100] ;

void PrintArray(int *a, int n)
{
    for (int i = 0; i < n; ++i)
        cout << a[i] << "";
    cout << endl ;
}

bool IsValid(int lastIndex, int value)
{
    for (int i = 0; i < lastIndex; i++)
    {
        if (buffer[i] >= value)
            return false;
    }
    return true;
}

void Select(int t, int n, int m)
{
    if (t == m)
        PrintArray(buffer, m);
    else
    {
        for (int i = 1; i <= n; i++)
        {
            buffer[t] = i;
            if (IsValid(t, i))
                Select(t + 1, n, m);
        }
    }
}

// 合并兩個(gè)有序數(shù)組
void Merge(int *a, int *b, int *c, int n)
{
    int i = 0 ;
    int j = 0 ;
    int k = 0 ;

    while (i < n && j < n)
    {
        if (a[i] < b[j])// 如果a的元素小，則插入a中元素到c
        {
            c[k++] = a[i] ;
            ++i ;
        }
        else if (a[i] == b[j])// 如果a和b元素相等，則插入二者皆可，這里插入a
        {
            c[k++] = a[i] ;
            ++i ;
            ++j ;
        }
        else // a[i] > b[j] // 如果b中元素小，則插入b中元素到c
        {
            c[k++] = b[j] ;
            ++j ;
        }
    }

    if (i == n) // 若a遍歷完畢，處理b中剩下的元素
    {
        for (int m = j; m < n; ++m)
            c[k++] = b[m] ;
    }
    else//j == n, 若b遍歷完畢，處理a中剩下的元素
    {
        for (int m = i; m < n; ++m)
            c[k++] = a[m] ;
    }
}

void Arrange(int* a, int n)
{
    int k = n -1 ;
    for (int i = n -1; i >=0; --i)
    {
        if (a[i] !=0)
        {
            if (a[k] ==0)
            {
                a[k] = a[i] ;
                a[i] =0 ;
            }
            --k ;
        }
    }
}

bool SameSign(int a, int b)
{
    if (a * b > 0)
        return true;
    else
        return false;
}

// 找出一個(gè)非遞減序整數(shù)序列中絕對(duì)值最小的數(shù)
int MinimumAbsoluteValue(int* a, int n)
{
    // Only one number in array
    if (n ==1)
    {
        return a[0] ;
    }

    // All numbers in array have the same sign
    if (SameSign(a[0], a[n -1]))
    {
        return a[0] >=0? a[0] : a[n -1] ;
    }

    // Binary search
    int l =0 ;
    int r = n -1 ;

    while(l < r)
    {
        if (l +1== r)
        {
            return abs(a[l]) < abs(a[r]) ? a[l] : a[r] ;
        }

        int m = (l + r) /2 ;

        if (SameSign(a[m], a[r]))
        {
            r = m -1;
            continue;
        }
        if (SameSign(a[l], a[m]))
        {
            l = m +1 ;
            continue;
        }
    }
}

// 找出一個(gè)非遞減序整數(shù)序列中絕對(duì)值最小的數(shù)
int MinimumAbsoluteValue(int* a, int n)
{
    // Only one number in array
    if (n ==1)
    {
        return a[0] ;
    }

    // All numbers in array have the same sign
    if (SameSign(a[0], a[n -1]))
    {
        return a[0] >=0? a[0] : a[n -1] ;
    }

    // Binary search
    int l =0 ;
    int r = n -1 ;

    while(l < r)
    {
        if (l + 1 == r)
        {
            return abs(a[l]) < abs(a[r]) ? a[l] : a[r] ;
        }

        int m = (l + r) /2 ;

        if (SameSign(a[m], a[r]))
        {
            r = m;
            continue;
        }
        else
        {
            l = m ;
            continue;
        }
    }
}