Anton and currency you all know------贪心

Berland, 2016. The exchange rate of currency you all know against the burle has increased so much that to simplify the calculations, its fractional part was neglected and the exchange rate is now assumed to be an integer.

Reliable sources have informed the financier Anton of some information about the exchange rate of currency you all know against the burle for tomorrow. Now Anton knows that tomorrow the exchange rate will be an even number, which can be obtained from the present rate by swapping exactly two distinct digits in it. Of all the possible values that meet these conditions, the exchange rate for tomorrow will be the maximum possible. It is guaranteed that today the exchange rate is an odd positive integer n. Help Anton to determine the exchange rate of currency you all know for tomorrow!

Input
The first line contains an odd positive integer n — the exchange rate of currency you all know for today. The length of number n’s representation is within range from 2 to 105, inclusive. The representation of n doesn’t contain any leading zeroes.

Output
If the information about tomorrow’s exchange rate is inconsistent, that is, there is no integer that meets the condition, print  - 1.

Otherwise, print the exchange rate of currency you all know against the burle for tomorrow. This should be the maximum possible number of those that are even and that are obtained from today’s exchange rate by swapping exactly two digits. Exchange rate representation should not contain leading zeroes.

Examples
Input
527
Output
572

Input
4573
Output
3574

Input
1357997531
Output
-1

题意

一个奇数n，交换其中不同的两位，使其变为偶数，求这个数最大是多少

思路

思路很简单，读完题就差不多出了，又造了几组数据试了一下，发现可行。我是这么考虑的，交换的话，肯定是优先考虑高位，同时还要保证数变大。所以就从高位向低位遍历，找出偶数，如果它比最后一位小就交换，如果一直到最后一个偶数都不比最后一位小的话，那就和最后一个偶数交换

代码

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int N = 1e5 + 10;
int main()
{
	char s[N];
	int a[N];
	cin >> s ;
	int n = strlen(s);
	int k = 0;
	int t;
	for (int i = 0; i < n; i++)
	{
		if (s[i] % 2 == 1)
		{
			a[i] = 1;
			k++;//奇数个数
		}
		else
		{
			a[i] = 0;//标记偶数
			t = i;
		}
	}
	if (k == n) 
	{
		cout << "-1" << endl;
		return 0;
	}
	for (int i = 0; i < n - 1; i++)
	{
		if (a[i] == 0 && s[i] < s[n - 1])//判断偶数与最后一位的大小
		{
			swap(s[i], s[n - 1]);
			break;
		}
		else if (i == t)swap(s[i], s[n - 1]);//偶数都大于等于最后一位
	}
	for (int i = 0; i < n; i++)
		cout << s[i];
	return 0;
}