题意: On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others. Seating in the class looks like a rectangle, where n rows with m pupils in each. The teacher asks pupils in the following or…

C. Unfair Poll time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others. Seat…

C. Unfair Poll time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others. Seat…

C. Unfair Poll time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others. Seat…

Unfair Poll 题意:一共有n排同学每排同学有m个人, 老师问问题有一个顺序, 先从第一排开始问,问完第一排的所有同学之后,再问第2排的,对于所有排的访问顺序为 1,2,3……n-1,n,n-1,n-2,……,2,1,2,然后每次访问到新的一排先要问完这一排的所有人才会往下一(目标)排走. 题解:先声明我们开一个数组来记录这一排被询问的总次数,先将k  /= m, 这个代表的是完全访问的次数,即一整排m位同学都问完有几次,如果  完全访问的次数< n, 我们就将前几排全访问次数的人都加上…

http://codeforces.com/problemset/problem/758/C C. Unfair Poll time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output On the Literature lesson Sergei noticed an awful injustice, it seems that some…

time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output On the Literature lesson Sergei noticed an awful injustice, it seems that some students are asked more often than others. Seating in the class lo…

http://codeforces.com/contest/758/problem/C 需要一个能够找到任意一个位置的步数的方法,就能解决三个问题. 预处理出one(row, col)表示第一次经过这个点,需要的步数. sec(row, col),thr(row, col)分别是第二.三步. 那么,循环节就会产生,记录步数为X, Y, Z,那么,第一二步的差距,和第二三步的差距,分别代表了从下走上去和从上走下来. 那么第4次,就会和第一二步的差距相等 就比如是5.17.23,那么下一步就是35.…

http://codeforces.com/problemset/problem/758/C 题意:教室里有n列m排,老师上课点名从第一列第一排开始往后点,直到点到第一列第m排,就从第二列第一排开始点,当点完第n列的名之后,接着点第n-1列的名.以此类推,就是从列上来看的话:1,2,3,4,……,n,n-1,n-2,……,1 ,2,…….这样的顺序点名.老师上课总共点k次名,问该课堂最多可以点同一个同学多少次,最少可以点同一个同学多少次,点了位置为(x,y)的同学多少次名. 思路:一遇到这种题目…

题意: 老师点名,顺序是1 -- n -- 1 排为一个循环,每列为1 -- m的顺序, 问点到最多次数和最少次数的人的次数以及(x,y)被点的次数. 分析: 由于点名有循环,故可先判断出每一个循环每个人被点名的次数,再乘以循环数,为答案一部分. 最后一个循环结束后k还有余数,从(1,1)暴力模拟,因为n*m才10000, 再加上前面的,就能得出答案. 注意 n=1 需要特判. #include <bits/stdc++.h> using namespace std; #define LL l…