A. Scarborough Fair
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Are you going to Scarborough Fair?

Parsley, sage, rosemary and thyme.

Remember me to one who lives there.

He once was the true love of mine.

Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there.

Willem asks his friend, Grick for directions, Grick helped them, and gave them a task.

Although the girl wants to help, Willem insists on doing it by himself.

Grick gave Willem a string of length n.

Willem needs to do m operations, each operation has four parameters l, r, c1, c2, which means that all symbols c1 in range [l, r] (froml-th to r-th, including l and r) are changed into c2. String is 1-indexed.

Grick wants to know the final string after all the m operations.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 100).

The second line contains a string s of length n, consisting of lowercase English letters.

Each of the next m lines contains four parameters l, r, c1, c2 (1 ≤ l ≤ r ≤ nc1, c2 are lowercase English letters), separated by space.

Output

Output string s after performing m operations described above.

Examples
input
3 1
ioi
1 1 i n
output
noi
input
5 3
wxhak
3 3 h x
1 5 x a
1 3 w g
output
gaaak
Note

For the second example:

After the first operation, the string is wxxak.

After the second operation, the string is waaak.

After the third operation, the string is gaaak.

【题意】:看例子。

【分析】:字符串下标从1开始。。scanf("%s", s+1);

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
int n,m;
char a[];
int L,R;
char c1,c2;
int main()
{
cin>>n>>m;
cin>>a;
for(int i=;i<=m;i++)
{
cin>>L>>R>>c1>>c2;
for(int i=L-;i<=R-;i++)
{
if(a[i]==c1)
{
a[i]=c2;
}
}
}
cout<<a<<endl;
}

注意字符串的 下标 从1开始!

Codeforces Round #449 (Div. 2) A. Scarborough Fair【多次区间修改字符串】的更多相关文章

  1. Codeforces Round #449 (Div. 2)-897A.Scarborough Fair(字符替换水题) 897B.Chtholly's request(处理前一半) 897C.Nephren gives a riddle(递归)

    A. Scarborough Fair time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  2. Codeforces Round #179 (Div. 1) A. Greg and Array 离线区间修改

    A. Greg and Array Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/295/pro ...

  3. Codeforces Round #449 (Div. 2)

    Codeforces Round #449 (Div. 2) https://codeforces.com/contest/897 A #include<bits/stdc++.h> us ...

  4. Codeforces Round #449 (Div. 2)ABCD

    又掉分了0 0. A. Scarborough Fair time limit per test 2 seconds memory limit per test 256 megabytes input ...

  5. Codeforces Round #449 Div. 2 A B C (暂时)

    A. Scarborough Fair 题意 对给定的长度为\(n\)的字符串进行\(m\)次操作,每次将一段区间内的某一个字符替换成另一个字符. 思路 直接模拟 Code #include < ...

  6. Codeforces Round #336 (Div. 2)【A.思维,暴力,B.字符串,暴搜,前缀和,C.暴力,D,区间dp,E,字符串,数学】

    A. Saitama Destroys Hotel time limit per test:1 second memory limit per test:256 megabytes input:sta ...

  7. Codeforces Round #390 (Div. 2) D. Fedor and coupons(区间最大交集+优先队列)

    http://codeforces.com/contest/754/problem/D 题意: 给定几组区间,找k组区间,使得它们的公共交集最大. 思路: 在k组区间中,它们的公共交集=k组区间中右端 ...

  8. Codeforces Round #526 (Div. 2) E. The Fair Nut and Strings

    E. The Fair Nut and Strings 题目链接:https://codeforces.com/contest/1084/problem/E 题意: 输入n,k,k代表一共有长度为n的 ...

  9. Codeforces Round #526 (Div. 2) D. The Fair Nut and the Best Path

    D. The Fair Nut and the Best Path 题目链接:https://codeforces.com/contest/1084/problem/D 题意: 给出一棵树,走不重复的 ...

随机推荐

  1. KVO的底层实现原理?如何取消系统默认的KVO并手动触发?

    KVO是基于runtime机制实现的 当某个类的属性对象第一次被观察时,系统就会在运行期动态地创建该类的一个派生类(该类的子类),在这个派生类中重写基类中任何被观察属性的setter 方法.派生类在被 ...

  2. python学习笔记十:异常

    一.语法 #!/usr/bin/python filename='hello' #try except finally demo try: open('abc.txt') print hello ex ...

  3. .netCore 反射 :Could not load file or assembly 系统找不到指定文件

    “System.IO.FileNotFoundException:“Could not load file or assembly 'ClassLibrary2, Culture=neutral, P ...

  4. cloud-init介绍及源码解读

    https://zhuanlan.zhihu.com/p/27664869 知乎大神写的  

  5. diskimage-builder element

    root阶段 创建或修改初始根文件系统内容. 这是添加替代分销支持的地方,还是建立在现有图像上的自定义. 只有一个元素可以一次使用它,除非特别注意不要盲目覆盖,而是适应其他元素提取的上下文. -cac ...

  6. Java面试准备十六:数据库——MySQL性能优化

    2017年04月20日 13:09:43 阅读数:6837 这里只是为了记录,由于自身水平实在不怎么样,难免错误百出,有错的地方还望大家多多指出,谢谢. 来自MySQL性能优化的最佳20+经验 为查询 ...

  7. 重复造轮子系列--字符串处理(C语言)

    这些字符代码是以前写的,源于很久很久以前的一个VC++项目,在当时的部门编程比赛里因为用了项目代码的xsplit函数,万万没想到,那个做了几年的项目里面居然有坑..xsplit函数居然不能split连 ...

  8. Oracle 数据库导出时 EXP-00008;ORA-00904

    问题是客户端和服务器端版本问题,我本地是11g,而服务器端是10g. 规则1:低版本的exp/imp可以连接到高版本(或同版本)的数据库服务器,但高版本的exp/imp不能连接到低版本的数据库服务器. ...

  9. 将MSHFlexGrid1中记录导出为Excel

    1.添加引用Microsoft Excel 14.0 Object Library 2.编写代码部分 Private Sub Output_Click() Dim i As Integer '定义变量 ...

  10. nginx 匹配路由分发php和golang

    大概这么个形式,可以走通 server { listen ; server_name localhost; root "E:/wwwroot180/public"; # 匹配指定路 ...