B. Anton and currency you all know
time limit per test
0.5 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Sample test(s)
input
527
output
572
input
4573
output
3574
input
1357997531
output
-1
题意：给一个长度不超过100000的奇数，要求交换任意两位上的数字，使它成为一个最大的偶数。
思路：字符串读入，记录下最后一位数上的奇数a，从前往后扫，找到第一个比a小的偶数，将它和a交换就是结果，若没有找到比a小的偶数，就将a和最后一个偶数交换。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 100005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
typedef long long ll;
using namespace std;
char str[maxn];
int main()
{
char ch;
while (~scanf("%s",str))
{
int len=strlen(str);
int heven=0; //标记是否有偶数
int minn=INF;
int pos; //记录第一个比最末尾奇数小的偶数的位置
int first=1;
int posmo; //记录最靠后的偶数的位置
int last=str[len-1]-'0';
for (int i=0;i<len;i++)
{
int a=str[i]-'0';
if (a%2==0)
{
posmo=i;
heven=1;
if (first&&a<last)
{
pos=i;
first=0;
break;
}
}
}
if (heven==0)
printf("-1\n");
else
{
if (!first)
{
ch=str[len-1];
str[len-1]=str[pos];
str[pos]=ch;
}
else
{
ch=str[len-1];
str[len-1]=str[posmo];
str[posmo]=ch;
}
printf("%s\n",str);
}
}
return 0;
}