| # | Problem | Pass Rate (passed user / total user) |
|---|---|---|
| 10840 | Check Palindrome |
|
| 10839 | Equivalent relation |
|
Description
Palindrome is a string that is identical to its reverse, like "level" or "aba". Check whether a given string is a palindrome or not.
Note that
1. This problem involves three files.
- function.h: Function definition of isPalindrome.
- function.c: Function describe of isPalindrome.
- main.c: A driver program to test your implementation.
You will be provided with main.c and function.h, and asked to implement function.c.
2. For OJ submission:
Step 1. Submit only your function.c into the submission block. (Please choose c compiler)
Step 2. Check the results and debug your program if necessary.
function.h
#ifndef FUNCTION_H #define FUNCTION_H int isPalindrome(char *start, char *end); #endif
main.c
#include <stdio.h>
#include <string.h>
#include "function.h"
int main()
{
char str[5000];
while (EOF != scanf("%s", str))
{
char *start = str;
char *end = start + strlen(str) - 1;
if (isPalindrome(start, end))
{
printf("Yes\n");
}
else
{
printf("No\n");
}
}
}
Input
The input consists of multiple lines. Each line contains a string.
The length of each string is less than 5000. The number of test case is less than 100.
Output
For each test case, output "Yes" if it's a palindrome or "No" if it's not a palindrome in a line.
Sample Input Download
Sample Output Download
Partial Judge Code
10840.cPartial Judge Header
10840.hTags
Discuss
Description
There are N integer pointers, indexed from 0 to N-1 (N<100). Each pointer initially points to an integer of value 0.
There are two kinds of instructions. The instruction “S n k” is used to set the integer, which pointer n points to, to be k. For example, S 1 10 means that the integer that pointer 1 points to is set to 10. And the instruction “P n m” means that pointer n points to the integer that pointer m points. For example, P 2 1 means that pointer 2 points to the integer that pointer 1 points to. After P 2 1, pointer 2 and pointer 1 point to the same integer, which is pointed by pointer 1.
Note that you don't have to change all the pointers if one pointer changes its target. The following table is an example. The instructions are P 1 2 and then P 2 3. You do not have to change the target of pointer 1.
|
instruction |
Description |
|
P 1 2 |
Pointer 1 points to the integer that pointer 2 points to.
|
|
P 2 3 |
Pointer 2 points to the integer that pointer 3 points to. And you don’t have to change the target of pointer 1. |
Finally, output all the values that pointers 0 to N-1 point to in order.
Note that
1. This problem involves three files.
- function.h: Function definition of execInstruct.
- function.c: Function describe of execInstruct.
- main.c: A driver program to test your implementation.
You will be provided with main.c and function.h, and asked to implement function.c.
2. For OJ submission:
Step 1. Submit only your function.c into the submission block. (Please choose c compiler)
Step 2. Check the results and debug your program if necessary.
Hints:
main.c
#include <stdio.h>
#include "function.h"
#define SIZE 100
int main() {
int *ptrArr[SIZE];
int dataArr[SIZE] = {0};
char inst;
int dataNum, instNum;
int param1, param2;
int i;
/* input */
scanf("%d %d", &dataNum, &instNum);
/* initialize the ptrArr */
for (i = 0; i < dataNum; i++)
ptrArr[i] = &dataArr[i];
for (i = 0; i < instNum; i++) {
scanf(" %c %d %d", &inst, ¶m1, ¶m2);
execInst(ptrArr, inst, param1, param2);
}
/* output */
for (i = 0; i < dataNum - 1; i++) {
printf("%d ", *ptrArr[i]);
}
printf("%d", *ptrArr[i]);
return 0;
}
function.h
#ifndef FUNCTION_H
#define FUNCTION_H
void execInst(int *ptrArr[], char inst, int param1, int param2);
#endif
Input
The first line contains two positive X and Y. X indicates the size of data. Y indicates that there are Y instructions needed to be done.
The next Y lines contain the instructions.
Output
All the values that pointers 0 to pointer N-1 point to in order. Each value is seperated by a blank ' '.
# Note that there is no '\n' at the end of the output.