JianZhiHao @ 2019-07-28 11:12:48
using namespace std; int a, b[100], c[100], k = 0, l; char h,m; int re(int x) { x = x - 26; if (x > 122) { re(x); } return x; } int main() { scanf("%d", &a); m=getchar(); for (int i = 97; i <= 122; i++) { b[k] = i; k++; } k = 0; while (1) { h = getchar(); if (h == '\n') break; l = h + a; if (l > 122) { l=re(l); } printf("%c",l);
}
return 0;
}
by 赫敏·东方延绪 @ 2019-07-28 11:16:13
希望更丰富的展现?使用Markdown
by 斗神_君莫笑 @ 2019-07-28 11:17:32
@JianZhiHao 找找上面那一行功能,有一个叫发送代码
by JianZhiHao @ 2019-07-28 11:22:31
#include <iostream>
using namespace std;
int a, b[100], c[100], k = 0, l;
char h,m;
int re(int x)
{
x = x - 26;
if (x > 122)
{
re(x);
}
return x;
}
int main()
{
scanf("%d", &a);
m=getchar();
for (int i = 97; i <= 122; i++)
{
b[k] = i;
k++;
}
k = 0;
while (1)
{
h = getchar();
if (h == '\n')
break;
l = h + a;
if (l > 122)
{
l=re(l);
}
printf("%c",l);
}
return 0;
}
by JianZhiHao @ 2019-07-28 11:23:05
@JianZhiHao 源码在此,谢谢!
by 人间失格 @ 2019-07-28 11:30:36
@JianZhiHao 应该是读入一行字符串,把每个字母转换成int后+N%26再转换成字符串吧
by JianZhiHao @ 2019-07-28 12:01:52
@人间失格 解决了 thanks!
by 人间失格 @ 2019-07-28 12:29:07
@JianZhiHao 不用谢