对于 n,求出所有小于等于n的 三次方数并放入数组中;对数组中的元素从大到小遍历,第一个回文数就是答案 #include<bits/stdc++.h>using namespacestd;#defineendl'\n'#defineint long longtypedeflonglongll;typedefpair<int,int> pii;constdoubleeps =1e-10;constintN =1e5+10;constintINF =1e16;constll...
AtCoder Beginner Contest 343 A - Wrong Answer (abc343 A) 题目大意 给定a,ba,b,输出cc,使得a+b≠ca+b≠c 解题思路 从00开始枚举cc的取值即可。 神奇的代码 #include<bits/stdc++.h> usingnamespacestd; usingLL =longlong; intmain(void){ ios::sync_with_stdio(false); cin.tie(0); cout.tie...
int>pii;constdoubleeps=1e-10;constintN=1e5+10;constintINF=1e16;constllmod=1e9+7;intis_(intx){inta=x,b=0;while(x){b=b*10+x%10;x/=10;}if(a==b)return1;return0;}voidsolve(){intn;cin>>n;vector<int>a;for(inti=1;i*i*i<=n;i++)a.push_...
AtCoder Beginner Contest 344A - SpoilerQuestion删除两个 | 之间的字符Solution按照题意模拟即可 Code#include <bits/stdc++.h> using namespace std; int main() { string s; cin >> s; string p1,…
We will hold AtCoder Beginner Contest 343.Contest URL: https://atcoder.jp/contests/abc343Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20240302T2100&p1=248Duration: 100 minutesWriter: yuto1115, cn449, evima, leaf1415Tester: Nyaan, math957963...
·AtCoder Beginner Contest (ABC) 这是最频繁且最简单的入门赛,通常情况下每月至少举行2次。2019年4月27日(含)之前,每场比赛共4题,时长100分钟,满分1000分且Rating超过1199的选手不计Rating值。自2019年5月19日起改版升级为6道题目,时长不变,满分2100分且Rating值超过1999的选手不计Rating值。改版之后比赛质量...
AtCoder Beginner Contest 361 | 完成度 [6 / 7] https://www.bilibili.com/video/BV1nZ421u7LG/ AtCoder Beginner Contest ABC-361-A https://www.bilibili.com/video/BV1tJbkewEx3/ AtCoder Beginner Contest ABC-361-B https://www.bilibili.com/video/BV1M7bkeYEh1/ AtCoder Beginner Contest ...
Σ 大赛——AtCoder Beginner Contest 353 https://www.bilibili.com/video/BV1kt421u7XK/ https://www.bilibili.com/video/BV1vf42127tH/ AtCoder Beginner Contest 353 实况(A~E) https://www.bilibili.com/video/BV1Gs421N748/ https://www.bilibili.com/video/BV19E42137MQ/ AtCoder Beginner Cont...
AtCoder Beginner Contest 043 题解 https://atcoder.jp/contests/abc043 A - Children and Candies (ABC Edit) 题目大意: 求\(1+2+\ldots+n\)。 答案:\(\frac{n \cdot (n+1)}{2}\)。 示例程序: #include <bits/stdc++.h>...
AtCoder Beginner Contest 043题解(ABCD) 传送门 A - Children and Candies (ABC Edit) 题意:求 ∑ i = 1 n i \sum\limits_{i=1}^n i i=1∑ni 思路:签到题,直接按照公式输出 n ( n + 1 ) 2 \dfrac{n(n+1)}{2} 2n(n+1)。