351681578wdp @ 2019-04-23 23:56:20

        std::cout <<char( (a[i]-'a'+num)%26+'a');
}
return 0;

by liuzitong @ 2019-04-24 07:58:04

因为WA==wonderful answer (完美的答案)

by SparrowHe @ 2019-04-24 10:17:05

@liuzitong WA == Wonderful Allkilled

by zhangqingbin125 @ 2019-04-25 15:52:05

include<iostream>

include<cstdio>

using namespace std;

int main()

{

int a;

cin>>a;

string b;

cin>>b;

int len=b.length();

for(int i=0;i<len;i++)

{
    cout<<char((b[i]-'a'+a)%26+'a');

}
cout<<endl;
return 0; 

}

by 大船靠岸 @ 2019-05-22 19:02:19

希望更丰富的展现？使用Markdown

#include <stdio.h>
#include <string.h>
#include <cstdio>
#include <iostream>
int main() {
    int num;
    std::cin >> num;
    char a[50];
    std::cin.get();
    gets_s(a);
    for (int i = 0; i < strlen(a); i++) {
        std::cout << char((a[i] - 'a' + num) % 26 + 'a');
    }
    return 0;
}