Digital collectible card games have become very popular recently. So Vova decided to try one of these. Vova has n cards in his collection. Each of these cards is characterised by its power pi, magic number ci and level li. Vova wants to build a deck…

After several latest reforms many tourists are planning to visit Berland, and Berland people understood that it's an opportunity to earn money and changed their jobs to attract tourists. Petya, for example, left the IT corporation he had been working…

Vasya has an array a consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the seco…

http://codeforces.com/problemset/problem/348/A A. Mafia time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output One day n friends gathered together to play "Mafia". During each round of the g…

Problem A Lucky Year 题目传送门[here] 题目大意是说,只有一个数字非零的数是幸运的,给出一个数,求下一个幸运的数是多少. 这个幸运的数不是最高位的数字都是零,于是只跟最高位有关,只保留最高位,然后加一求差就是答案. Code /** * Codeforces * Problem#808A * Accepted * Time:15ms * Memory:0k */ #include<iostream> #include<cstdio> #include<…

传送门 题意 将n个数划分为两块,最多改变一个数的位置, 问能否使两块和相等 分析 因为我们最多只能移动一个数x,那么要么将该数往前移动,要么往后移动,一开始处理不需要移动的情况 那么遍历sum[i] 如果往前移动,sum[k]+(sum[i]-sum[i-1])=sum[n]/2,k∈[1,i-1] 如果往后移动,sum[k]-(sum[i]-sum[i-1])=sum[n]/2,k∈[i+1,n] 一开始我没有考虑往后移动 时间复杂度\(O(nlog(n))\) trick 代码 #incl…

这题1<<M为255,可以logN二分答案后,N*M扫一遍表把N行数据转化为一个小于等于255的数字,再255^2检验答案(比扫一遍表复杂度低),复杂度约为N*M*logN #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ][]; ][]; int n,m; int ans,ans2; ],pre[]; int zhuangya(int x,int y){ ;i<=m;++i)…

D. String Game time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty…

D. String Game Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, because she is too young. Therefore, her brother Sergey always helps her. Sergey gives Nastya th…

Educational Codeforces Round 21  A. Lucky Year 个位数直接输出\(1\) 否则,假设\(n\)十进制最高位的值为\(s\),答案就是\(s-(n\mod s)\) view code #pragma GCC optimize("O3") #pragma GCC optimize("Ofast,no-stack-protector") #include<bits/stdc++.h> using namespac…