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  • Codeforces Round #449 (Div. 2) A. Scarborough Fair【多次区间修改字符串】

    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[150];
    int L,R;
    char c1,c2;
    int main()
    {
            cin>>n>>m;
            cin>>a;
            for(int i=1;i<=m;i++)
            {
              cin>>L>>R>>c1>>c2;
              for(int i=L-1;i<=R-1;i++)
              {
                  if(a[i]==c1)
                  {
                      a[i]=c2;
                  }
              }
            }
            cout<<a<<endl;
    }
    注意字符串的 下标 从1开始!
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  • 原文地址:https://www.cnblogs.com/Roni-i/p/7956260.html
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