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  • Codeforces Round #587 (Div. 3) A. Prefixes

    链接:

    https://codeforces.com/contest/1216/problem/A

    题意:

    Nikolay got a string s of even length n, which consists only of lowercase Latin letters 'a' and 'b'. Its positions are numbered from 1 to n.

    He wants to modify his string so that every its prefix of even length has an equal amount of letters 'a' and 'b'. To achieve that, Nikolay can perform the following operation arbitrary number of times (possibly, zero): choose some position in his string and replace the letter on this position with the other letter (i.e. replace 'a' with 'b' or replace 'b' with 'a'). Nikolay can use no letters except 'a' and 'b'.

    The prefix of string s of length l (1≤l≤n) is a string s[1..l].

    For example, for the string s="abba" there are two prefixes of the even length. The first is s[1…2]="ab" and the second s[1…4]="abba". Both of them have the same number of 'a' and 'b'.

    Your task is to calculate the minimum number of operations Nikolay has to perform with the string s to modify it so that every its prefix of even length has an equal amount of letters 'a' and 'b'.

    思路:

    从开头开始每两个字符需要不同, 搞一下就行了.

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    const int MAXN = 2e5+10;
    
    int n;
    char s[MAXN];
    
    int main()
    {
        cin >> n;
        cin >> s;
        int op = 0, a = 0, b = 0;
        for (int i = 0;i < n;i+=2)
        {
            if (s[i] != s[i+1])
                continue;
            if (s[i] == 'a')
                s[i] = 'b';
            else
                s[i] = 'a';
            op++;
        }
        cout << op << endl;
        cout << s << endl;
    
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/YDDDD/p/11622744.html
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