HDOJ(HDU) 2137 circumgyrate the string(此题用Java-AC不过！坑)

0
0
0
1. 云栖社区>
2. 博客>
3. 正文

## HDOJ(HDU) 2137 circumgyrate the string(此题用Java-AC不过！坑)

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.

Input
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.

Output
For each case, print the circumgrated string.

Sample Input
asdfass 7

``````Sample Output
a
s
d
f
a
s
s``````

Java这个代码完全没问题的，至少我还没找到错哪了，如果有找到的，求告知，谢谢。

WA的Java代码：

``````package cn.hncu.acm;

import java.util.Scanner;

public class P2137 {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()) {
String str = sc.next();
int n = sc.nextInt();
n %= 8;
if (n < 0)
n += 8;
int t = str.length();
if (n == 0)
System.out.println(str);
else if (n == 4) {
for (int i = t - 1; i >= 0; i--)
System.out.print(str.charAt(i));
System.out.println();
} else if (n == 1) {
for (int i = t - 1; i >= 0; i--) {
for (int j = i; j > 0; j--)
System.out.print(" ");
System.out.println(str.charAt(i));
}
} else if (n == 2) {
for (int i = t - 1; i >= 0; i--) {
for (int j = 0; j < (t / 2); j++)
System.out.print(" ");
System.out.println(str.charAt(i));
}
} else if (n == 3) {
for (int i = t - 1; i >= 0; i--) {
for (int j = t - 1; j > i; j--)
System.out.print(" ");
System.out.println(str.charAt(i));
}
} else if (n == 5) {
for (int i = 0; i < t; i++) {
for (int j = i + 1; j < t; j++)
System.out.print(" ");
System.out.println(str.charAt(i));
}
} else if (n == 6) {
for (int i = 0; i < t; i++) {
for (int j = 0; j < t / 2; j++)
System.out.print(" ");
System.out.println(str.charAt(i));
}
} else if (n == 7) {
for (int i = 0; i < t; i++) {
for (int j = 0; j < i; j++)
System.out.print(" ");
System.out.println(str.charAt(i));
}
}
}
}
}
``````

AC的C语言代码：

``````#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char name[100];
int n;

int main()
{
while( scanf("%s",name) !=EOF ){
int t = strlen(name);
scanf("%d",&n);
n %= 8;
if( n < 0 )
n += 8;
if( n == 0 )
printf("%s\n",name);
else if( n==4 ){
for( int i=t-1 ; i>=0 ; i-- )
printf("%c",name[i]);
printf("\n");
}
else if( n == 1 ){
for( int i=t-1; i>=0 ; i-- ){
for( int j=i ; j>0 ; j-- )
printf(" ");
printf("%c\n",name[i]);
}
}
else if( n == 2 ){
for( int i=t-1 ; i>=0 ; i-- ){
for( int j=0 ; j<(t/2) ; j++ )
printf(" ");
printf("%c\n",name[i]);
}
}
else if( n == 3 ){
for( int i=t-1 ; i>=0 ; i-- ){
for( int j=t-1 ; j>i ; j-- )
printf(" ");
printf("%c",name[i]);
printf("\n");
}
}
else if( n == 5 ){
for( int i=0 ; i<t ; i++ ){
for( int j=i+1 ; j<t ; j++ )
printf(" ");
printf("%c\n",name[i]);
}
}
else if( n == 6 ){
for( int i=0 ; i<t ; i++ ){
for( int j=0 ; j<t/2 ; j++ )
printf(" ");
printf("%c\n",name[i]);
}
}
else if( n == 7 ){
for( int i=0 ; i<t ; i++ ){
for( int j=0 ; j<i ; j++ )
printf(" ");
printf("%c\n",name[i]);
}
}
}
return 0;
}  ``````

+ 关注

corcosa 13153人浏览