算法模板总结

Posted 2021-03-22 shuitiangong

排序：

1.冒泡排序(什么？你也是冒泡人？):

void bubble_Sort(int* arr, int n) {
    for (int i = 0; i<n-1; ++i)
        for (int j = 0; j<n-i-1; ++j)
            if (arr[j] > arr[j+1])
                swap(arr[j], arr[j+1]);
}

2.快速排序:

void quickSort(int* arr, int low, int high) {
    if (low >= high)
        return;
    int kase = low;
    for (int i = low; i<high; i++) 
        if (arr[i] < arr[high])
            swap(arr[kase++], arr[i]);
    swap(arr[kase], arr[high]);
    quickSort(arr, low, kase-1);
    quickSort(arr, kase+1, high);
}

3.归并排序：

void mergeSort(int arr[], int temp[], int low, int high) {
    if (low >= high)
        return;
    int len = high-low, mid = len/2 + low;
    int start1 = low, end1 = mid, start2 = mid+1, end2 = high;
    mergeSort(arr, temp, start1, end1);
    mergeSort(arr, temp, start2, end2);
    int index = low;
    while(start1 <= end1 && start2 <= end2) 
        temp[index++] = arr[start1] <= arr[start2] ? arr[start1++] : arr[start2++];
    while(start1 <= end1)
        temp[index++] = arr[start1++];
    while(start2 <= end2)
        temp[index++] = arr[start2++];
    for (int i = low; i<=high; i++)
        arr[i] = temp[i];
}

4.待续....

快速幂&&快速乘：

1.快速幂：

ll quickPow(ll x, ll y) {
    if (!y)
        return 1;
    ll res = quickPow(x, y >> 1);
    res = res*res%MOD;
    if (y&1)
        res = res*x%MOD;
    return res;
}

2.快速乘：

ll solve(ll n, ll m) {
    if (!n || !m)
        return 0;
    ll res = solve(n, m>>1);
    res = res<<1 % MOD;
    if (m&1)
        res = (res+n) % MOD;
    return res;
}

 高精度：

　1.加法

void sum() {
        int len1 = strlen(num1);
        int len2 = strlen(num2);
        reverse(num1, num1+len1);
        reverse(num2, num2+len2);
        int kase = 0;
        while(kase < len1 && kase < len2) {
            res[kase+1] = (res[kase] + num1[kase] + num2[kase] - ‘0‘*2)/10;
            res[kase] = (res[kase] + (num1[kase] + num2[kase] - ‘0‘*2))%10 + ‘0‘;
            ++kase;
        }
        while(kase < len1) {
            res[kase+1] = (res[kase] + num1[kase] - ‘0‘)/10;
            res[kase] = (res[kase] + num1[kase] - ‘0‘)%10 + ‘0‘;
            ++kase;
        }
        while(kase < len2) {
            res[kase+1] = (res[kase] + num2[kase] - ‘0‘)/10;
            res[kase] = (res[kase] + num2[kase] - ‘0‘)%10 + ‘0‘;
            ++kase;
        }
        res[kase] += ‘0‘;
        for (int i = kase; i>=0; i--) {
            if (res[kase] == ‘0‘)
                continue;
            cout << res[i];
        }
        cout << endl;
        fill(num1, num1+maxn, 0);
        fill(num2, num2+maxn, 0);
        fill(res, res+maxn, 0);
}

　尺取法：