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.
The first line contains two integers l and r (2 ≤ l ≤ r ≤ 109).
Print single integer, the integer that appears maximum number of times in the divisors.
If there are multiple answers, print any of them.
19 29
2
3 6
3
Definition of a divisor: https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html
The first example: from 19 to 29 these numbers are divisible by 2: {20, 22, 24, 26, 28}.
The second example: from 3 to 6 these numbers are divisible by 3: {3, 6}.
哈哈，数学推导，试几个样例直接拿代码试啊，毕竟DIV2 A轻轻松松可以水过的，写这些题就是想标记下，希望自己以后可以拿数论推一下
#include <bits/stdc++.h> using namespace std; int main() { int a,b; scanf("%d%d",&a,&b); if(a==b) printf("%d\n",a); else if(b-a==3&&b%3==0) printf("%d\n",3); else printf("2\n"); return 0; }
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.
The first line contains single integer n (1 ≤ n ≤ 2·105) — the length of the string.
Print the string that satisfies all the constraints.
If there are multiple answers, print any of them.
2
aa
3
bba
A palindrome is a sequence of characters which reads the same backward and forward.
B题脑洞题，类似于模拟，挺简单的
#include<stdio.h> char a[5]="aabb"; int main() { int n; scanf("%d",&n); int c=0; for(int i=0;i<n;i++) { printf("%c",a[c%4]); c++; } return 0; }
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.
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of schools.
Print single integer: the minimum cost of tickets needed to visit all schools.
2
0
10
4
In the first example we can buy a ticket between the schools that costs .
继续脑洞啊，肯定是让首尾连两个结合啊，然后再往中间走，然后不就是（n-1）/2了
#include <stdio.h> int main(){ int n; scanf("%d",&n); printf("%d\n",n-1>>1); return 0;}
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.
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Print the minimum number of steps modulo 109 + 7.
ab
1
aab
3
The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa".
有点像做过的PAT这种题，就是有只有这两种字符，不是b就是a啊，多推一会得到公式就好，有点dp的感觉
#include <bits/stdc++.h> using namespace std; const int N=1e9+7; int main() { string s; cin>>s; int ans=0; int t=0; for(int i=s.length()-1;i>=0;i--) { if(s[i]=='b') t=(t+1)%N; else { ans=(ans+t)%N; t=(t*2)%N; } } cout<<ans<<endl; return 0; }