A. Easy Math Time Limit: 2000ms Memory Limit: 65536KB Given n integers a1,a2,-,an, check if the sum of their square root a1−−√+a2−−√+⋯+an−−√ is a integer. Input The input consists of multiple tests. For each test: The first line contains 1 integer n…

J. Usoperanto Time Limit: 8000ms Memory Limit: 256000KB Usoperanto is an artificial spoken language designed and regulated by Usoperanto Academy. The academy is now in study to establish Strict Usoperanto, a variation of the language intended for for…

C. Censor Time Limit: 2000ms Memory Limit: 65536KB frog is now a editor to censor so-called sensitive words (敏感词). She has a long text p. Her job is relatively simple – just to find the first occurence of sensitive word w and remove it. frog repeats…

B. Carries Time Limit: 1000ms Memory Limit: 65536KB frog has n integers a1,a2,-,an, and she wants to add them pairwise. Unfortunately, frog is somehow afraid of carries (进位). She defines \emph{hardness} h(x,y) for adding x and y the number of carries…

I. Travel Time Limit: 3000ms Memory Limit: 65536KB The country frog lives in has n towns which are conveniently numbered by 1,2,-,n. Among n(n−1)/2 pairs of towns, m of them are connected by bidirectional highway, which needs a minutes to travel. The…

J. Right turn Time Limit: 1000ms Memory Limit: 65536KB frog is trapped in a maze. The maze is infinitely large and divided into grids. It also consists of n obstacles, where the i-th obstacle lies in grid (xi,yi). frog is initially in grid (0,0), hea…

E. Rectangle Time Limit: 1000ms Memory Limit: 65536KB 64-bit integer IO format: %lld Java class name: Main Submit Status frog has a piece of paper divided into n rows and m columns. Today, she would like to draw a rectangle whose perimeter is not gre…

题目链接 https://acm.bnu.edu.cn/v3/contest_show.php?cid=8506#problem/A problem description As we know, the NTU Final PK contest usually tends to be pretty hard. Many teams got frustrated when participating NTU Final PK contest. So I decide to make the fi…

题目链接 题目大意就是这个,先找出下标的循环节,再快速幂对20160519取余就行了. 找出下标循环节: #include <cstdio> #include <iostream> using namespace std; int main() { ,a=,b=; for(;;i++) { int t=b; b=(a+b)%; a=t%; &&b==) { printf("%d\n",i); break; } } ; } AC代码: #inclu…

题目链接 把左括号看成A右括号看成B,推一下就行了.好久之前写的,推到最后发现是一个有规律的序列. #include <bits/stdc++.h> using namespace std; typedef long long ll; int main() { ll n; while(scanf("%lld",&n)!=EOF) { ll cnt=,sum=; ;; i++) { if(sum>=n) break; cnt++; sum=sum+cnt; //…