链接：

https://codeforces.com/contest/1251/problem/C

题意：

You are given a huge integer a consisting of n digits (n is between 1 and 3?105, inclusive). It may contain leading zeros.

You can swap two digits on adjacent (neighboring) positions if the swapping digits are of different parity (that is, they have different remainders when divided by 2).

For example, if a=032867235 you can get the following integers in a single operation:

302867235 if you swap the first and the second digits;
023867235 if you swap the second and the third digits;
032876235 if you swap the fifth and the sixth digits;
032862735 if you swap the sixth and the seventh digits;
032867325 if you swap the seventh and the eighth digits.
Note, that you can‘t swap digits on positions 2 and 4 because the positions are not adjacent. Also, you can‘t swap digits on positions 3 and 4 because the digits have the same parity.

You can perform any number (possibly, zero) of such operations.

Find the minimum integer you can obtain.

Note that the resulting integer also may contain leading zeros.

思路：

双指针奇偶数，奇偶比较大小即可。

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;

string s;

int main()
{
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while(t--)
    {
        cin >> s;
        int p1 = -1, p2 = -1;
        int len = s.length();
        for (int i = 0;i < len;i++)
        {
            if (p1 == -1 && ((int)s[i]-'0')%2 == 0)
                p1 = i;
            if (p2 == -1 && ((int)s[i]-'0')%2 == 1)
                p2 = i;
        }
        while(p1 < len && p2 < len && p1 != -1 && p2 != -1)
        {
            if ((int)(s[p1]-'0') < (int)(s[p2]-'0'))
            {
                printf("%c", s[p1]);
                p1++;
                while(p1 < len && (int)(s[p1]-'0')%2 == 1)
                    p1++;
            }
            else
            {
                printf("%c", s[p2]);
                p2++;
                while(p2 < len && (int)(s[p2]-'0')%2 == 0)
                    p2++;
            }
        }
        if (p1 < len && p1 != -1)
        {
            for (int i = p1;i < len;i++)
            {
                if ((int)(s[i]-'0')%2 == 0)
                    printf("%c", s[i]);
            }
        }
        else if (p2 < len && p2 != -1)
        {
            for (int i = p2;i < len;i++)
            {
                if ((int)(s[i]-'0')%2 == 1)
                    printf("%c", s[i]);
            }
        }
        puts("");
    }

    return 0;
}

原文地址：https://www.cnblogs.com/YDDDD/p/11779189.html