题意:给你 n 段区间,而且还是不相交的,然后你只能向左扩展左端点,或者向右扩展右端点,然后扩展最少的步数让整数总数能够整除 k. 析:很简单么,只要在记录算一下数量,然后再算出 k 的倍数差多少就行. 代码如下: #include <bits/stdc++.h> using namespace std; typedef long long LL; const int maxn = 500 + 5; const int INF = 0x3f3f3f3f; const int dr[] = {0…

题意:给定一个字符,让你用前 k 个字符把它排成 n 长度,相邻的字符不能相等,并且把字典序最小. 析:其实很简单么,我们只要多循环ab,就行,最后再把剩下的放上,要注意k为1的时候. 代码如下: #include <bits/stdc++.h> using namespace std; typedef long long LL; const int maxn = 10000 + 5; const int INF = 0x3f3f3f3f; const int dr[] = {0, 0, 1,…

L. Ministry of Truth Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100187/problem/K Description Andrey works in the Ministry of Truth. His work is changing articles in newspapers and magazines so that they praise the Party an…

B. Archer Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/312/problem/B Description SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The…

A. Fashion in Berland 题目连接: http://www.codeforces.com/contest/691/problem/A Description According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if…

A. Opponents 题目连接: http://www.codeforces.com/contest/688/problem/A Description Arya has n opponents in the school. Each day he will fight with all opponents who are present this day. His opponents have some fighting plan that guarantees they will win…

后天考试,今天做题,我真佩服自己... 这次又只A俩水题... orz各路神犇... 话说这次模拟题挺多... 半个多小时把前面俩水题做完,然后卡C,和往常一样,题目看懂做不出来... A: 算是模拟吧,反正看懂题目后很可耻的生硬水果,没被hack我觉得非常开心... 由于要求两个人至少一个没跳过,画图后发现很偷懒的方法. 虽然跟样例不同,但过了... #include <cstdio> int main() { int n, m; scanf("%d%d", &n…

对于这道水题本人觉得应该应用贪心算法来解这道题: 下面就贴出本人的代码吧: #include<cstdio> #include<iostream> using namespace std; ],b[]; int main(void) { int n; ; ,sum2 = ; ;i<=;++i){ scanf("%d",&a[i]); sum1 += a[i]; } ;i<=;++i){ scanf("%d",&b[…

C. Four Segments 题目连接: http://codeforces.com/contest/14/problem/C Description Several months later Alex finally got his brother Bob's creation by post. And now, in his turn, Alex wants to boast about something to his brother. He thought for a while,…

题目链接:http://codeforces.com/problemset/problem/289/B 题目意思:给出一个 n 行 m 列的矩阵和数值 d .通过对矩阵里面的数进行 + d 或者 - d 的操作,是否可以使矩阵上的所有数相等.可以的话输出最少的操作步数,否则输出 -1. 由于矩阵的排列对处理没什么影响,因此不需要用到二维数组存储.接着把矩阵中所有的数从小到大进行排序,要想算出最少的步数,很容易想到应该最中间的数(中位数)靠拢.最关键的是如何判断不能通过对矩阵中的数进行处理使得所有…