nefuoj-1041：字符串变形记

description

<span style="font-family:SimSun;font-size:18px;">Problem Description Give you a string, just circumgyrate. The number N means you just circumgyrate the string N times, and each time you circumgyrate the string for 45 degree anticlockwise. </span>

input

<span style="font-family:SimSun;font-size:18px;">In each case there is string and a integer N. And the length of the string is always odd, so the center of the string will not be changed, and the string is always horizontal at the beginning. The length of the string will not exceed 80, so we can see the complete result on the screen. </span>

output

<span style="font-family:SimSun;font-size:18px;">For each case, print the circumgrated string.</span>

sample_input

<span style="font-family:SimSun;font-size:18px;">asdfass 7</span>

sample_output

<span style="font-family:SimSun;font-size:18px;">a s d f a s s</span>

题解：模拟，纯粹的模拟，而且还得挨个模拟，敲着好累啊。。。

code：

#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

int main()

{

char s[105];

int n;

while(cin>>s)

{

cin>>n;

n%=8;

int len=strlen(s);

if(n==0)

{

puts(s);

continue;

}

if(n==7)

{

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

{

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

putchar(' ');

printf("%c\n",s[i]);

}

continue;

}

if(n==6)

{

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

{

printf("%c\n",s[i]);

}

continue;

}

if(n==5)

{

for(int i=len-1; i>=0; i--)

{

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

putchar(' ');

printf("%c\n",s[len-i-1]);

}

continue;

}

if(n==4)

{

for(int i=len-1; i>=0; i--)

putchar(s[i]);

putchar('\n');

continue;

}

if(n==3)

{

for(int i=len-1; i>=0; i--)

{

for(int j=0; j<len-i-1; j++)

putchar(' ');

printf("%c\n",s[i]);

}

continue;

}

if(n==2)

{

for(int i=len-1; i>=0; i--)

{

printf("%c\n",s[i]);

}

continue;

}

if(n==1)

{

for(int i=len-1; i>=0; i--)

{

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

putchar(' ');

printf("%c\n",s[i]);

}

continue;

}

}

return 0;

}

下边的code是学校给的答案，说起来用switch语句看起来的确会舒服一点，但是字符数并没有少多少，，。，，

code2：

#include <cstdio>

#include <cstring>

#include <iostream>

using namespace std;

int main()

{

char str[100];

int n,i,j;

while(scanf("%s%d",str,&n)!=-1)

{

if(n>=0) n%=8;

else

{

n%=8;n+=8;n%=8;

}

int len=strlen(str);

switch(n)

{

case 0:

printf("%s\n",str);

break;

case 1:

for(i=len-1;i>=0;i--)

for(j=0;j<=i;j++)

if(j==i) printf("%c\n",str[i]);

else printf(" ");

break;

case 2:

for(i=len-1;i>=0;i--)

for(j=0;j<=len/2;j++)

if(j==len/2) printf("%c\n",str[i]);

else printf(" ");

break;

case 3:

for(i=len-1;i>=0;i--)

for(j=0;j<=len-1-i;j++)

if(j==len-1-i) printf("%c\n",str[i]);

else printf(" ");

break;

case 4:

for(i=len-1;i>=0;i--)

printf("%c",str[i]);

printf("\n");

break;

case 5:

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

for(j=0;j<=len-1-i;j++)

if(j==len-1-i) printf("%c\n",str[i]);

else printf(" ");

break;

case 6:

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

for(j=0;j<=len/2;j++)

if(j==len/2) printf("%c\n",str[i]);

else printf(" ");

break;

case 7:

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

for(j=0;j<=i;j++)

if(j==i) printf("%c\n",str[i]);

else printf(" ");

break;

}

}

return 0;

}