题目地址:http://codeforces.com/contest/435/problem/C /* 题意:给一组公式,一组数据,计算得到一系列的坐标点,画出折线图:) 模拟题:蛮恶心的,不过也简单,依据公式得知折线一定是一上一下的,只要把两个相邻的坐标之间的折线填充就好 注意:坐标有可能为负数,所以移动值y初始化为1000,最后要倒过来输出 详细解释:http://blog.csdn.net/zhb1997/article/details/27877783 */ #include <cstd…

http://codeforces.com/contest/435/problem/C…

B. Pasha Maximizes time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Pasha has a positive integer a without leading zeroes. Today he decided that the number is too small and he should make it…

链接:http://codeforces.com/contest/435/problem/A   A. Queue on Bus Stop time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output It's that time of the year when the Russians flood their countryside su…

题目链接: http://codeforces.com/contest/435/problem/D D. Special Grid time limit per test:4 secondsmemory limit per test:256 megabytes 问题描述 You are given an n × m grid, some of its nodes are black, the others are white. Moreover, it's not an ordinary gri…

D.E还是很难的.....C不想多说什么... A:提意:给出每一组人的个数,以及一次车载容量,求出最少需要多少次才能载走所有的人. water: http://codeforces.com/contest/435/submission/6741997 B:提意:给出一个数,最多可以交换两个数k次,并且每次只能交换相邻的数,求最大的数是多少. water: 很水的贪心,直接从后往前扫描k位,把最大的数转移到最高位就行了.只不过晚上杀了一晚的后缀数组,木了,搞了半天.. http://codefo…

C. Cardiogram time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output In this problem, your task is to use ASCII graphics to paint a cardiogram. A cardiogram is a polyline with the following corner…

C. Cardiogram time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output In this problem, your task is to use ASCII graphics to paint a cardiogram. A cardiogram is a polyline with the following corner…

水题 #include <iostream> #include <vector> #include <algorithm> using namespace std; int main(){ vector<int> a(4); cin >> a[0] >> a[1]>>a[2]>>a[3]; string str; cin >> str; int res = 0; for(int i = 0 ; i…

看到题目的时候,以为类似插入排序,比较第i个元素和第i-1个元素, 如果第i个元素比第i-1个元素小,则不交换 如果第i个元素比第i-1个元素大,则交换第i个元素和第i-1个元素 继续比较第i-1个元素与前一个元素,直到前一个元素大为止 交换元素次大于等于k则停止 但对测试用例 1234 3 则出现问题,如果按照上述方法得到答案为3214, 但正确答案是4321,直接将第4个元素往前交换 故在上面基础上改进得 当扫描第i个元素时,则要取出[i,i+k+1)之间的最大元素,然后将最大元素往前交换,…