B. Om Nom and Dark Park Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/526/problem/B Description Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living a…

Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to hel…

D. Om Nom and Necklace time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output One day Om Nom found a thread with n beads of different colors. He decided to cut the first several beads from this th…

C. Om Nom and Candies Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/526/problem/C Description A sweet little monster Om Nom loves candies very much. One day he found himself in a rather tricky situation that required him…

传送门 C. Om Nom and Candies time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A sweet little monster Om Nom loves candies very much. One day he found himself in a rather tricky situation that r…

题目描述: 有一天,欧姆诺姆发现了一串长度为n的宝石串,上面有五颜六色的宝石.他决定摘取前面若干个宝石来做成一个漂亮的项链. 他对漂亮的项链是这样定义的,现在有一条项链S,当S=A+B+A+B+A+...+A+B+A的时候是漂亮的,这儿A,B是一些宝石串,“+”表示连接操作.S中有k+1个A和k个B组成.A和B可能是空串. 现在给出宝石串,问怎么切前几个才能得到一个漂亮的宝石项链.他切下来之后不会改变宝石的顺序. 样例解释: 在这个样例中前6个可以组成漂亮的串( A="", B=&qu…

拖了好久的题解,想想还是补一下吧. A. King of Thieves 直接枚举起点和5个点之间的间距,进行判断即可. #include <bits/stdc++.h> using namespace std; ]; int main() { //freopen("in.txt", "r", stdin); int n; bool ans = false; scanf("%d%s", &n, s); ; q < n &…

B. Om Nom and Dark Park 在满二叉树上的某些边上添加一些值.使得根节点到叶子节点的路径上的权值和都相等.求最少需要添加多少. 我们利用性质解题.   考察兄弟节点.由于他们从跟节点到父节点这路径是相同的,所以需要添加的值为  2*max(lch,rch)-lch-rch;  同时将max(lch,rch)累加到父节点. 思路是最重要的.有时候不是聪不聪明的问题,而是会不会思考的问题. #include <iostream> #include <cstdio>…

注意题目给的是一个nxm的park,设元素为aij,元素aij 有4种可能U(上移),D(下移),L(左移),R(右移) 假设第i行第j列元素aij(注意元素的索引是从0开始的) 当aij为D时,此时spiders一直往下移动不可能与Om Nom相遇 当aij为U时,此时spiders向上移动时此时Norm与spider初始位置之间距离是i+1,注意索引是从0开始的 如果i为偶数,i+1为奇数则两者必能在i/2位置相遇 如果i为奇数,当Norm在i/2,sprider在i/2+1,移动时,Nor…

Om Nom and Spiders time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Om Nom really likes candies and doesn't like spiders as they frequently steal candies. One day Om Nom fancied a walk in a…