c to c 2b 2b converter online

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Carla
26 Nov 2018
1
2#ifndef CMAC_H
3#define CMAC_H
4void AES_CMAC(unsigned char *key, unsigned char *input, int length,
5	unsigned char *mac);
6
7#endif
Chaima
24 Aug 2017
1#include <stdio.h>
2
3int main()
4{
5   
6   // C++ program to find sum of two large numbers.
7#include<bits/stdc++.h>
8using namespace std;
9
10// Function for finding sum of larger numbers
11string findSum(string str1, string str2)
12{
13	// Before proceeding further, make sure length
14	// of str2 is larger.
15	if (str1.length() > str2.length())
16		swap(str1, str2);
17
18	// Take an empty string for storing result
19	string str = "";
20
21	// Calculate length of both string
22	int n1 = str1.length(), n2 = str2.length();
23
24	// Reverse both of strings
25	reverse(str1.begin(), str1.end());
26	reverse(str2.begin(), str2.end());
27
28	int carry = 0;
29	for (int i=0; i<n1; i++)
30	{
31		// Do school mathematics, compute sum of
32		// current digits and carry
33		int sum = ((str1[i]-'0')+(str2[i]-'0')+carry);
34		str.push_back(sum%10 + '0');
35
36		// Calculate carry for next step
37		carry = sum/10;
38	}
39
40	// Add remaining digits of larger number
41	for (int i=n1; i<n2; i++)
42	{
43		int sum = ((str2[i]-'0')+carry);
44		str.push_back(sum%10 + '0');
45		carry = sum/10;
46	}
47
48	// Add remaining carry
49	if (carry)
50		str.push_back(carry+'0');
51
52	// reverse resultant string
53	reverse(str.begin(), str.end());
54
55	return str;
56}
57
58// Driver code
59int main()
60{
61	string str1 = "12";
62	string str2 = "198111";
63	cout << findSum(str1, str2);
64	return 0;
65}
66 printf("Hello World");
67
68    return 0;
69}
70
Katrin
18 Jun 2018
1// C++ program to generate all possible
2// valid IP addresses from given string
3#include <bits/stdc++.h>
4using namespace std;
5
6// Function checks whether IP digits
7// are valid or not.
8int is_valid(string ip)
9{
10	// Splitting by "."
11	vector<string> ips;
12	string ex = "";
13	for (int i = 0; i < ip.size(); i++) {
14		if (ip[i] == '.') {
15			ips.push_back(ex);
16			ex = "";
17		}
18		else {
19			ex = ex + ip[i];
20		}
21	}
22	ips.push_back(ex);
23
24	// Checking for the corner cases
25	// cout << ip << endl;
26	for (int i = 0; i < ips.size(); i++) {
27		// cout << ips[i] <<endl;
28		if (ips[i].length() > 3
29			|| stoi(ips[i]) < 0
30			|| stoi(ips[i]) > 255)
31			return 0;
32
33		if (ips[i].length() > 1
34			&& stoi(ips[i]) == 0)
35			return 0;
36
37		if (ips[i].length() > 1
38			&& stoi(ips[i]) != 0
39			&& ips[i][0] == '0')
40			return 0;
41	}
42	return 1;
43}
44
45// Function converts string to IP address
46void convert(string ip)
47{
48	int l = ip.length();
49
50	// Check for string size
51	if (l > 12 || l < 4) {
52		cout << "Not Valid IP Address";
53	}
54
55	string check = ip;
56	vector<string> ans;
57
58	// Generating different combinations.
59	for (int i = 1; i < l - 2; i++) {
60		for (int j = i + 1; j < l - 1; j++) {
61			for (int k = j + 1; k < l; k++) {
62				check = check.substr(0, k) + "."
63						+ check.substr(k, l - k + 2);
64				check
65					= check.substr(0, j) + "."
66					+ check.substr(j, l - j + 3);
67				check
68					= check.substr(0, i) + "."
69					+ check.substr(i, l - i + 4);
70
71				// cout<< check <<endl;
72				// Check for the validity of combination
73				if (is_valid(check)) {
74					ans.push_back(check);
75					std::cout << check << '\n';
76				}
77				check = ip;
78			}
79		}
80	}
81}
82
83// Driver code
84int main()
85{
86	string A = "25525511135";
87	string B = "25505011535";
88
89	convert(A);
90	convert(B);
91
92	return 0;
93}
94
95// This code is contributed by Harshit
96
Angelo
09 Apr 2016
1    string word;
Basil
04 Aug 2016
1123456789101112131415161718192021222324252627#include <iostream>#include <bits/stdc++.h>using namespace std;class ABS{    public:        void abs(int arr[], int n){    sort(arr, arr+n);    int c=0,i;    for (i=n-1;i-2>=0;i--){        if (arr[i-2]+arr[i-1]>arr[i]){            c=1;            break;        }    }    (c) ? cout<<arr[i-2]<<" "<<arr[i-1]<<" "<<arr[i]<<endl : cout<<-1<<endl;}};int main(){    ABS abx;    int arr[] = { 5,4,3,1,2};    int n = sizeof(arr) / sizeof(arr[0]);    abx.abs(arr, n);    return 0;}X
Tevin
22 Aug 2018
1123456789101112#include <stdio.h>#if !defined (MESSAGE)#define MESSAGE "You wish!"#endifint main(void){cout << "Hello World" << endl;return 0;}X
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