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  • codeforces gym 100971 K Palindromization 思路

    题目链接:http://codeforces.com/gym/100971/problem/K

    K. Palindromization
    time limit per test
    2.0 s
    memory limit per test
    256 MB
    input
    standard input
    output
    standard output

    Mihahim has a string s. He wants to delete exactly one character from it so that the resulting string would be a palindrome. Determine if he can do it, and if he can, what character should be deleted.

    Input

    The input contains a string s of length (2 ≤ |s| ≤ 200000), consisting of lowercase Latin letters.

    Output

    If the solution exists, output «YES» (without quotes) in the first line. Then in the second line output a single integer x — the number of the character that should be removed from s so that the resulting string would be a palindrome. The characters in the string are numbered from 1. If there are several possible solutions, output any of them.

    If the solution doesn't exist, output «NO» (without quotes).

    Examples
    input
    evertree
    output
    YES
    2
    input
    emerald
    output
    NO
    input
    aa
    output
    YES
    2

     题意:给你一个字符串,删除一个字符,是否形成回文,如果有,输出删除的位置;

    思路:对于删除一个字符以后,这个字符串的对称轴的位置只有两个,所有枚举对称轴,复杂度O(n);

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<algorithm>
    #include<cstring>
    #include<vector>
    using namespace std;
    #define LL long long
    const int N=2e5+100,M=1e6+10,inf=1e9+10;
    const LL INF=1e18+10,mod=19260817;
    
    char a[N];
    int checkl(int l,int r,int n,int L)
    {
        while(r<=n)
        {
            while(a[l]!=a[r])
            {
                if(L!=1)return 0;
                L=l;
                l--;
            }
            l--;r++;
        }
        return L;
    }
    int checkr(int l,int r,int n,int R)
    {
        while(l>=1)
        {
            while(a[l]!=a[r])
            {
                if(R!=n)return 0;
                else R=r;
                r++;
            }
            l--;r++;
        }
        return R;
    }
    int main()
    {
        scanf("%s",a+1);
        int n=strlen(a+1);
        if(n%2==0)
        {
            int ans=checkl(n/2,n/2+2,n,1);
            if(ans)return 0*printf("YES
    %d
    ",ans);
            ans=checkr(n/2-1,n/2+1,n,n);
            if(ans)return 0*printf("YES
    %d
    ",ans);
            puts("NO");
        }
        else
        {
            int ans=checkl(n/2+1,n/2+2,n,1);
            if(ans)return 0*printf("YES
    %d
    ",ans);
            ans=checkr(n/2,n/2+1,n,n);
            if(ans)return 0*printf("YES
    %d
    ",ans);
            puts("NO
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jhz033/p/7441705.html
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