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. 两个category方法相同调用哪个

    Category扩展,它是对一个类进行功能的扩展.在项目的开发过程中,在不断的迭代开发过程中,我们的类也不可避免的要根据需求来增加新的功能,而这个时候很多的人可能会新建一个子类,然后在子类中去增加我们 ...

  2. Python 3基础教程6-for循环语句

    本文介绍另外一种循环语句,for循环,直接看例子. 用for实现打印1到9的数字. 方法一:写入一个列表,然后遍历列表 # 这里介绍 for循环# 打印1到9 exampleList = [1,2,3 ...

  3. navmesh自动寻路

    一个导航网格(也就是Navmesh)是世界几何体简化的表示法,被游戏代理用于在世界中进行导航.通常,代理(agent )有一个目标,或一个目的地,它试图找到一个路径,然后沿路径导航到达目标.这个过程被 ...

  4. vue 搜索匹配

    computed: { broSeachData: function() { var browesData = this.browesData, searchVal = this.searchVal; ...

  5. rownum浅谈(一)

    只要做web开发,几乎没有不需要分页查询的,在oracle中,rownum就是用来进行处理分页的. 1.rownum是oracle对结果集返回的一个伪列,也就是说是先查询完结果之后再加上的一个虚列,相 ...

  6. poj1273 网络流入门题 dinic算法解决,可作模板使用

    Drainage Ditches Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 62078   Accepted: 2384 ...

  7. Swift中由找不到removeAll(where:)方法引起的连锁反应(上)

    核心代码 section.removeAll {baseRow in if let habitRow = baseRow as? HabitRow{ let idxPath = habitRow.in ...

  8. HDU 5690 矩阵快速幂

    All X Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submi ...

  9. HDU 1054树形DP入门

    Strategic Game Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) ...

  10. linux查看日志的方法

    linux查看日志文件内容命令tail.cat.tac.head.echo tail -f test.log你会看到屏幕不断有内容被打印出来. 这时候中断第一个进程Ctrl-C, ---------- ...