B. Marbles time limit per test0.5 seconds memory limit per test1024 megabytes Using marbles as a currency didn't go so well in Cubicônia. In an attempt to make it up to his friends after stealing their marbles, the Emperor decided to invite them to a game night in his palace. Of ...
Letaaandbbbe two arrays of lengthsnnandmm, respectively, with no elements in common. We can define a new arraymerge(a,b)merge(a,b)of lengthn+mn+mrecursively as follows: If one of the arrays is empty, the result is the other array. That is,merge(∅,b)=bmerge(∅,b)=bandmerg...
Codeforces #658 Problem B 题面 There are n piles of stones, where the i-th pile has ai stones. Two people play a game, where they take alternating turns removing stones. In a move, a player may remove a positive number of stones from the first non-empty pile (the pile with the ...
codeforces.com/contest/1201/problem/B B. Zero Array 一道思维题 因为每次操作对于sum会减少2,所以如果sum是奇数肯定不可以。另外,如果最大值大于前面n-1个数的总和,也一定不可以。其他情况都可以。 证明: 如果最后一个数等于前面n-1个数的总和,那显然可以。 最后一个数小于前面n-1个数的总和,通过每次操作,...
https://codeforces.com/contest/1090/problem/B Examples standard input The most famous characters of Pushkin’s works are Onegin \cite{onegin}, Dubrovsky \cite{dubrovsky} and Tsar Saltan \cite{saltan}. \begin{thebibliography}{99} ...
CodeForces round 753 problem B Odd Grasshopper(奇怪的蚱蜢),题目翻译:Thegrasshopper(蚂蚱)islocatedonthenumericaxis(数轴)atthepointwithcoordinate(坐标)x0.一个蚱蜢在数轴上,初始坐标为x
codeforces Add solution for problem B - GCD Compression Sep 21, 2021 README.md README.md Add solution for problem E - Knapsack 2 Mar 26, 2022 submissions.json submissions.json Add solution for problem E - Knapsack 2 Mar 26, 2022 View all files Repository files navigation README Submissions...
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你的算法需要优化。其实不需要去找出每对匹配的i j，只要找出数量就可以了。我给段简单的代码，你看看应该就明白了。include <stdio.h> double table[256];char str[100001];void main(){ char *p = str;int i;double n=0;gets(str);while(*p)table[*(p++)]+=1;for (int i = 1; i...
从左到右有n个木块,编号为0-n-1,要求模拟以下4种操作(下面a和b都是木块编号)。 move a onto b: 把a和b上方的木块全部归位,然后把a摞到b上面。 move a over b: 把a上方的木块全部归位,然后把a及上面的木块整体摞到b上面。 pile a onto b: 把b上方的木块全部归位,然后把a及a上面的木块摞在b上面...