codefoces9001多名

Codeforces是一个知名的在线编程竞赛平台，它定期举办各种算法和技术挑战赛，包括个人赛、练习赛以及合作解题等。"Codeforces 9001名"很可能是指某个特定比赛或者等级的要求，比如达到Codeforces用户排名前9001位。在Codeforces上取得高排名需要具备扎实的编程基础、良好的解题技巧，以及经常参与训练和实战比赛来提升能力。如果你提到这个数字是为了参加比赛或提升自己的排名，那么持续学习、刷题并参加官方活动是很重要的。

相关问题

codefoces 1272D

Sure! Here's a possible solution for Codeforces problem 1272D:

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main() {
    int n;
    cin >> n;

    vector<int> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }

    sort(a.begin(), a.end());

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

    cout << count << endl;

    return 0;
}

This solution takes an input `n`, followed by `n` integers. It sorts the array and then iterates through it, checking for duplicates or decreasing values. If a duplicate is found, it increments the next element to make them different. If a decreasing value is found, it increments the next element to make it greater than the current one. Finally, it prints the total number of operations performed.

Please note that this is just one possible solution, and there may be other approaches as well.