2019-5-8训练

http://codeforces.com/problemset/problem/1144/E

26进制，任意进制的做法

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
int k;
char aa[200010];
char bb[200010];
int a[200010];
int b[200010];
int zz[200010];
int ans[200010];
int main()
{
    cin >> k;
    for (int i = 1; i <= k; i++)
    {
        cin >> aa[i];
        a[i] = int(aa[i] - 'a');
    }
    for (int i = 1; i <= k; i++)
    {
        cin >> bb[i];
        b[i] = int(bb[i] - 'a');
    }
    for (int i = k; i >=1; i--)
    {
        if (b[i] >= a[i])
        {
            zz[i] = b[i] - a[i];
        }
        else
        {
            zz[i] = b[i] + 26 - a[i];
            b[i - 1]--;
        }
    }
    for (int i = 1; i <= k; i++)
    {
        if (zz[i] % 2 == 0)
        {
            ans[i] = zz[i] / 2;
        }
        else
        {
            ans[i] = (zz[i] - 1) / 2;
            zz[i + 1] = zz[i + 1] + 26;
        }
    }
    for (int i = k; i >= 1; i--)
    {
        if (ans[i] + a[i] >= 26)
        {
            ans[i] = ans[i] + a[i] - 26;
            ans[i - 1]++;
        }
        else
        {
            ans[i] = ans[i] + a[i];
        }
    }
    for (int i = 1; i <= k; i++)
    {
        printf("%c", char(ans[i]+'a'));
    }
    return 0;
}