比赛传送门 只能说当晚状态不佳吧,有点头疼感冒的症状.也跟脑子没转过来有关系,A题最后一步爆搜没能立即想出来,B题搜索没有用好STL,C题也因为前面两题弄崩了心态,最后,果然掉分了. A:简单数学 B:数学+排序 C:字符串搜索 A // https://codeforces.com/contest/1277/problem/A /* 题意: 给出一个数,求不大于该数的完美整数的个数(完美整数指全是一个数组成的数字,如:111, 333333, 444444, 9, 8888 ...) 分析:…

概述 切了 ABCE,Room83 第一 还行吧 A - Happy Birthday, Polycarp! 题解 显然这样的数不会很多. 于是可以通过构造法,直接求出 \([1,10^9]\) 内所有符合要求的数. \(\mathrm{Code}\) #include<bits/stdc++.h> using namespace std; int a[]={1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,77,88,99,111,222,333,444,555,666…

Happy Birthday, Polycarp! Make Them Odd As Simple as One and Two Let's Play the Words? Two Fairs Beautiful Rectangle Happy Birthday, Polycarp! \[ Time Limit: 1 s\quad Memory Limit: 256 MB \] 暴力枚举所有数字全为 \(i.i\in[1,9]\),然后暴力判断有多个全为 \(i\) 的数在 \(n\) 以内即可…

链接 签到题,求出位数,然后9*(位数-1)+ 从位数相同的全一开始加看能加几次的个数 #include<bits/stdc++.h> using namespace std; int main(int argc, char const *argv[]) { int t; int y; cin>>t; int ans = 0; while(t--) { cin>>y; ans = 0; int weishu = 0; int temp = y; while(temp&g…

A - Forgetting Things 题意:给 \(a,b\) 两个数字的开头数字(1~9),求使得等式 \(a=b-1\) 成立的一组 \(a,b\) ,无解输出-1. 题解:很显然只有 \(a=b\) 和 \(a=b-1\) 的时候有解,再加上一个从 \(9\) 越到 \(10\) 的就可以了.被样例的 \(199,200\) 误导了. #include<bits/stdc++.h> using namespace std; typedef long long ll; int mai…

链接: https://codeforces.com/contest/1241/problem/D 题意: You are given a sequence a1,a2,-,an, consisting of integers. You can apply the following operation to this sequence: choose some integer x and move all elements equal to x either to the beginning,…

链接: https://codeforces.com/contest/1247/problem/D 题意: You are given n positive integers a1,-,an, and an integer k≥2. Count the number of pairs i,j such that 1≤i<j≤n, and there exists an integer x such that ai⋅aj=xk. 思路: 对每个数质数拆分,记录质数的指数modk的值,map记录ve…

https://codeforces.com/contest/1241/problem/C You are an environmental activist at heart but the reality is harsh and you are just a cashier in a cinema. But you can still do something! You have n tickets to sell. The price of the i-th ticket is pi.…

链接: https://codeforces.com/contest/1241/problem/C 题意: You are an environmental activist at heart but the reality is harsh and you are just a cashier in a cinema. But you can still do something! You have n tickets to sell. The price of the i-th ticket…

A,有多个线段,求一条最短的线段长度,能过覆盖到所又线段,例如(2,4)和(5,6) 那么我们需要4 5连起来,长度为1,例如(2,10)(3,11),用(3,10) 思路:我们想一下如果题目说的是最长我们肯定是取最小x和最大的y连起来就完事. 但是要求长度最小又得覆盖,那么可以这样想,我们需要把最小的x不断右移到这条线段的y, 最大的左移到x,所以就是最大x-最小y完事 #include <bits/stdc++.h> using namespace std; #define ll long…