CodeForces 81A

一眼看上去,题目很长,其实就是去重,把相邻且相同的两个字符同时去掉,直到没有相邻且相同的字符为止

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <string>
 4 #include <cstring>
 5 #include <string.h>
 6 using namespace std;
 7 char s[200002];
 8 int main(){
 9     scanf("%s",s);
10     int len=strlen(s);
11     int j=1;
12     for(int i=1;i<len;i++){
13         if(s[i]!=s[j-1])s[j++]=s[i];             //当前的字符与新串中的最后一个字符不一样,就复制
14         else {
15             j--;s[j]=‘\0‘;               //当前的字符与新串中的最后一个字符一样把新串最后一个删除了
16         }
17     }
18     s[j]=‘\0‘;                        //把j后面的字符都不要了,printf("%s",s),会在遇到‘\0’,的时候结束输出
19     printf("%s\n",s);
20 }

CodeForces 81A

时间: 2024-10-16 03:07:38

CodeForces 81A的相关文章

CSU-ACM暑假集训基础组训练赛(4)解题报告

•Problem A SPOJ SUB_PROB   AC自动机 •题意: 给定一个长为M(M≤100000 )的文本串,和N(N≤1000)个长度不超过2000的模式串,问每个模式串是否在文本串中出现过? •几乎和周一课件上的第一个例题一模一样.. •把文本串丢到AC自动机里面去跑. •注意: •1.可能有两个相同的模式串(略坑吧.) •2.一个模式串可能是另一个模式串的后缀,即如果一个点的fail指针指向的点是一个“危险节点”,那么它本身也是一个“危险节点”. 1 #include <ios

【codeforces 718E】E. Matvey&#39;s Birthday

题目大意&链接: http://codeforces.com/problemset/problem/718/E 给一个长为n(n<=100 000)的只包含‘a’~‘h’8个字符的字符串s.两个位置i,j(i!=j)存在一条边,当且仅当|i-j|==1或s[i]==s[j].求这个无向图的直径,以及直径数量. 题解:  命题1:任意位置之间距离不会大于15. 证明:对于任意两个位置i,j之间,其所经过每种字符不会超过2个(因为相同字符会连边),所以i,j经过节点至多为16,也就意味着边数至多

Codeforces 124A - The number of positions

题目链接:http://codeforces.com/problemset/problem/124/A Petr stands in line of n people, but he doesn't know exactly which position he occupies. He can say that there are no less than a people standing in front of him and no more than b people standing b

Codeforces 841D Leha and another game about graph - 差分

Leha plays a computer game, where is on each level is given a connected graph with n vertices and m edges. Graph can contain multiple edges, but can not contain self loops. Each vertex has an integer di, which can be equal to 0, 1 or  - 1. To pass th

Codeforces Round #286 (Div. 1) A. Mr. Kitayuta, the Treasure Hunter DP

链接: http://codeforces.com/problemset/problem/506/A 题意: 给出30000个岛,有n个宝石分布在上面,第一步到d位置,每次走的距离与上一步的差距不大于1,问走完一路最多捡到多少块宝石. 题解: 容易想到DP,dp[i][j]表示到达 i 处,现在步长为 j 时最多收集到的财富,转移也不难,cnt[i]表示 i 处的财富. dp[i+step-1] = max(dp[i+step-1],dp[i][j]+cnt[i+step+1]) dp[i+st

Codeforces 772A Voltage Keepsake - 二分答案

You have n devices that you want to use simultaneously. The i-th device uses ai units of power per second. This usage is continuous. That is, in λ seconds, the device will use λ·ai units of power. The i-th device currently has bi units of power store

Educational Codeforces Round 21 G. Anthem of Berland(dp+kmp)

题目链接:Educational Codeforces Round 21 G. Anthem of Berland 题意: 给你两个字符串,第一个字符串包含问号,问号可以变成任意字符串. 问你第一个字符串最多包含多少个第二个字符串. 题解: 考虑dp[i][j],表示当前考虑到第一个串的第i位,已经匹配到第二个字符串的第j位. 这样的话复杂度为26*n*m*O(fail). fail可以用kmp进行预处理,将26个字母全部处理出来,这样复杂度就变成了26*n*m. 状态转移看代码(就是一个kmp

Codeforces Round #408 (Div. 2) B

Description Zane the wizard is going to perform a magic show shuffling the cups. There are n cups, numbered from 1 to n, placed along the x-axis on a table that has m holes on it. More precisely, cup i is on the table at the position x?=?i. The probl

Codeforces 617 E. XOR and Favorite Number

题目链接:http://codeforces.com/problemset/problem/617/E 一看这种区间查询的题目,考虑一下莫队. 如何${O(1)}$的修改和查询呢? 令${f(i,j)}$表示区间${\left [ l,r \right ]}$内数字的异或和. 那么:${f(l,r)=f(1,r)~~xor~~f(1,l-1)=k}$ 记一下前缀异或和即可维护. 1 #include<iostream> 2 #include<cstdio> 3 #include&l