这次CF感觉挺简单的，但是我思路不够宽

A. Fake NP

Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path.

You are given l and r. For all integers from l to r, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times.

Solve the problem to show that it's not a NP problem.

Input
The first line contains two integers l and r (2 ≤ l ≤ r ≤ 109).

Output
Print single integer, the integer that appears maximum number of times in the divisors.

If there are multiple answers, print any of them.

这题签到题，如果 l==r ，那么出现次数最多为1，如果 l!=r 那么出现次数最多的为0

int main() {
    int l, r;
    cin >> l >> r;
    if (l == r) {
        cout << l << endl;
    }
    else {
        cout << 2 << endl;
    }

    return 0;
}

B. 3-palindrome
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick.

He is too selfish, so for a given n he wants to obtain a string of n characters, each of which is either 'a', 'b' or 'c', with no palindromes of length 3 appearing in the string as a substring. For example, the strings "abc" and "abca" suit him, while the string "aba" doesn't. He also want the number of letters 'c' in his string to be as little as possible.

Input
The first line contains single integer n (1 ≤ n ≤ 2·105) — the length of the string.

Output
Print the string that satisfies all the constraints.

If there are multiple answers, print any of them.

这题题目要求输出满足条件的字符串，C经可能小，那么只用不断的输出aa，bb，即可

int main(){

    int n;
    cin >> n;
    string fuck = "aabb";
    for(int i = 0;i < n; i++){
        putchar(fuck[i%4]);
    }
    puts("");

    return 0;
}

C. Find Amir
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
A few years ago Sajjad left his school and register to another one due to security reasons. Now he wishes to find Amir, one of his schoolmates and good friends.

There are n schools numerated from 1 to n. One can travel between each pair of them, to do so, he needs to buy a ticket. The ticker between schools i and j costs  and can be used multiple times. Help Sajjad to find the minimum cost he needs to pay for tickets to visit all schools. He can start and finish in any school.

Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of schools.

Output
Print single integer: the minimum cost of tickets needed to visit all schools

题目要求输出最少的花费来经过全部学校，其中票可以复用。我们 先分析下，票的花费函数是这样的： Cost=(i+j)mod(n+1)

那么第一种情况就是 i=1，j=n ，这样花费为0，然后i，j不断的向n/2逼近，保证每次消耗都是最少的。
以 1，2，3 为例

Cost1=(1+3)mod(1+3)

Cost2=(3+2)mod(1+3)

Cost=Cost1+Cost2

消耗就为1

int main()
{
        int n;
        cin>>n;
        cout<<(n+1)/2-1<<endl;
}

D. Minimum number of steps
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.

The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.

Input
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.

Output
Print the minimum number of steps modulo 109 + 7.

这题是输出按照要求变换字符串所需的最小次数