CodeForces - 451C(模拟)

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 oddpositive 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 from2 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.

Example
Input
527
Output
572
Input
4573
Output
3574
Input
1357997531
Output
-1

题意:给你一个数n,可以任意交换任意两个数字,是这个数字变成偶数,并且要使这个数尽可能的大。

解:用字符串存一下这一串数字,然后判断给的这一串数字中是否存在偶数,若不存在直接输出-1,否则从最高位开始遍历字符串,找到第一个小于最后一位的偶数数字,然后交换这两个数字即可,若没有偶数数字小于最后一位,那么就把最后一位偶数数字跟最后那位数字交换即可。这样就可以保证所得的数是最大的了。详细请看代码:

#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long LL;
const int N=1e5+1;
char s[N];
int main()
{
    bool flag=true;
    scanf("%s",s);
    int len=strlen(s);
    for(int i=0; s[i]!='\0'; i++)
        if((s[i]-'0')%2==0) flag=false;//标记是否存在偶数
    if(flag) printf("-1\n");//若果不存在直接输出-1
    else
    {
        int j;
        flag=true;
        for(int i=0;s[i]!='\0';i++)
        {
            if((s[i]-'0')%2==0)
            {
                if((s[i]-'0')<(s[len-1]-'0'))//找小于最后一位数的偶数
                {
                    flag=false;
                    char t=s[len-1];
                    s[len-1]=s[i];
                    s[i]=t;
                    printf("%s\n",s);
                    break;
                }
                j=i;//如果没有偶数数字小于最后一位,记录一下最后一位偶数的下标
            }
        }
        if(flag)
        {
            char t=s[len-1];
            s[len-1]=s[j];
            s[j]=t;
            printf("%s\n",s);
        }
    }
}




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