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!
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.
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.
527
572
4573
3574
1357997531
-1
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define maxn 100005 char a[maxn]; int main() { int i,j; // freopen("in.txt","r",stdin); while(scanf("%s",&a)!=EOF){ i=0; j=strlen(a)-1; int flag=0; int flag1=0; while(i<=j){ if(a[j]>a[i]&&(a[i]-'0')%2==0){ //从前往后找第一个比最后一个小的数，交换，可以这样想像，前面的都比最后一个小，肯定和第一个数换能够得到最大值 swap(a[i],a[j]); flag=1; break; } i++; } if(flag==0){ j=strlen(a)-1; while(j>=0){ if(a[j]>a[strlen(a)-1]&&(a[j]-'0')%2==0){ //从后往前找第一个比最后一个小的数，交换，这样想，如果前面的都比最后一个大，要交换的那个数肯定越后越好。 swap(a[j],a[strlen(a)-1]); flag1=1; break; } j--; } } if(flag1==1||flag==1)puts(a); else puts("-1"); } }
B. Anton and currency you all know
原文地址：http://blog.csdn.net/u013497977/article/details/43229741
