which type of information could reasonably be expected to cause serious damage to national security if disclosed without authori
1 answer:
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Answer:
The answer is "64"
Explanation:
There has been 1 Private Key 1 Public Key for every person. The private key is used for other app messages. All else uses the Public Key to encrypt messages with that user. Those keys are numerically related. Even if you have 5 users, you will be given 5 Personal keys and 5 Public keys.
- All user will have a duplicate of other public keys, which indicates (n*n-1) copies of n public keys on different systems to make sure shared contact between all users, plus n private keys.
- The unique key count is 2n, with n^2 distributed private and public keys on different systems.
- 8 people, where 8 is a number, that is equal to n so,
To find out the duration of alex's movies, you can use the code in C++ to make this calculation, where you will only need the number of movies and their duration, as well:
<h3>Writing code in C++:</h3><em>#include<bits/stdc++.h></em>
<em>using namespace std;</em>
<em>struct Movie {</em>
<em> int timeBegin, duration, timeEnd;</em>
<em> bool operator<(const Movie& another) const {</em>
<em> return timeEnd < another.timeEnd;</em>
<em> }</em>
<em>};</em>
<em>struct Festival {</em>
<em> int count;</em>
<em> vector<Movie> movies;</em>
<em>};</em>
<em>Festival* initialize(int timeBegin[], int duration[], int count) {</em>
<em> Festival* filmFestival = new Festival;</em>
<em> filmFestival->count = count;</em>
<em> for (int i = 0; i < count; i++) {</em>
<em> Movie temp;</em>
<em> temp.timeBegin = timeBegin[i];</em>
<em> temp.duration = duration[i];</em>
<em> temp.timeEnd = timeBegin[i] + duration[i];</em>
<em> filmFestival->movies.push_back(temp);</em>
<em> }</em>
<em> return filmFestival;</em>
<em>}</em>
<em>int solve(Festival* fest) {</em>
<em> int res = 0;</em>
<em> sort(fest->movies.begin(), fest->movies.end());</em>
<em> int timeEnd = -1;</em>
<em> for (int i = 0; i < fest->count; i++) {</em>
<em> if (fest->movies[i].timeBegin >= timeEnd) {</em>
<em> res++;</em>
<em> timeEnd = fest->movies[i].timeEnd;</em>
<em> }</em>
<em> }</em>
<em> return res;</em>
<em>}</em>
<em>int main(int argc, char *argv[]) {</em>
<em>int timeBegin[] = {1, 3, 0, 5, 5, 8, 8};</em>
<em>int duration[] = {3, 2, 2, 4, 3, 2, 3};</em>
<em>Festival * fest;</em>
<em>fest = initialize(timeBegin,duration, 7);</em>
<em>cout << solve(fest) << endl;</em>
<em>return 0;</em>
<em>}</em>
See more about C Code at brainly.com/question/17544466
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