codeforces 508B Anton and currency you all know

B. Anton and currency you all know
time limit per test
0.5 seconds

memory limit per test
256 megabytes

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 ofcurrency 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 anodd 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 numbern's representation is within range from
2 to 105, inclusive. The representation ofn 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.

Sample test(s)

Input

527

Output

572

Input

4573

Output

3574

Input

1357997531

Output

-1

题目链接：http://codeforces.com/contest/508/problem/B

题目大意：给一个奇数，只交换其中两个，得到一个最大的偶数，若不存在输出－1

题目分析：从最低位往高位找，找最后一位小于第一位的偶数，若不存在小于第一位的偶数则找第一个出现的偶数，因为后面的偶数都比第一位大，交换了后面的与第一位得到的肯定不是最大的

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
char s[100005];

int main()
{
    int tmp = -1;
    scanf("%s", s);
    int len = strlen(s);
    for(int i = len - 2; i >= 0; i--)
        if((s[i] - '0') % 2 == 0)
            if(s[i] < s[len - 1] || tmp == -1)
                tmp = i;
    if(tmp == -1)
        printf("-1\n");
    else
    {
        swap(s[tmp], s[len - 1]);
        printf("%s\n", s);
    }
}