Description

Cheat Sheet

printf() and  scanf() format

printf("%d", n);
 

FORMAT  ARGUMENT TYPE

%d, %i  int           decimal

%u      unsigned int

%x      unsigned int  hexadecimal

%#x     unsigned int  hexadecimal with prefix 0x

%f      double  

%Lf     long double

%e, %E  double      scientific notation

%c      int         to print a character

%s      char *      string (character array ended with '\0')

%p      void *      print memory address

%g, %G  double      %f or %e depending on the length

scanf("%d", &n);
 

FORMAT  ARGUMENT TYPE

%d      int *       &n, store the input integer in n

%ld     long *

%lld    long long *

%u      unsigned int *

%f      float *     read float

%lf     double *    read double

%Lf     long double *   read long double

%c      char *      read 3 characters %3c

%s      char *      read a string until whitespace

%n      int *       with %s, to get string length

                   char a[100]; int len; 
                  scanf("%s%n", a, &len);
                  len will have the string length

 

Frequently used functions

#include <string.h>
char str[10];
scanf("%s", str);
to get the string length using strlen(str)
to compare two strings strcmp(str1, str2) ==0 if equal
to compare the first n chars of two strings strncmp(str1, str2, n) ==0 if equal
to copy str2 to str1 strcpy(str1, str2)
to copy the first n chars of str2 to str1 strncpy(str1,str2, n) remember to add '\0' to str1
#include <ctype.h>
isspace(ch), islower(ch), isupper(ch), isdigit(ch)
isalpha(ch), toupper(ch), tolower(ch)

To create a 5-by-5 two-dimensional array, we need to write

int a[5][5];

 

It will be indexed as follows:

 

 

 

How to read the following data?
1 2 3 4 5 e
#include <stdio.h>
int main(void)
{
    int x;
    while (scanf("%d", &x) == 1) {   
     printf("x=%d\n", x);
    }
    return 0;
}

How to read the following data?

2

L 5 2
D 5 3

#include <stdio.h>

int main(void)

{

   char ch;
   int i, n, row, col;

   scanf("%d", &n);

   for (i=0; i<n; i++) {

      while(getchar()!='\n');

      scanf("%c%d%d", &ch, &row, &col);

   }

   return 0;

}

 

Using for loops to print a two-dimensional array

   for(i = 0; i < row; i++) {
      for (j = 0; j < col; j++) {
         printf("%5d", A[i][j]);
      }
      printf("\n");
   } 

Using bubble sort to rearrange an array A

for (i = 0; i < n; i++) {
   for (j = 1; j < n; j++) {
      if (A[j] > A[j-1]) {
         /* swap A[j] A[j-1] */
      }
   }
}

operators:
!   &&    ||    ==     !=    +    -    *    /    %
>   <     >=    <=

How to avoid common errors and how to debug for OJ

1. Put the arrays in the 'global' area. Set their size bigger than required. Avoid using variable-length arrays (e.g. int arr[n];). Keep the array size fix (e.g., int arr[1000];).

2. After writing the code for reading input data, you may print out the data to check if your code reads them correctly. Do not proceed to write subsequent code before you confirm that.

3. If your program crashes, usually it is caused by memory related errors. Check the ranges of for-loops to see if your code attempts to read or write the elements out of the arrays’ boundary.

*(a+i) is equivalent to a[i]
(a+i) is equivalent to &a[i]

qsort :

you have to include <stdlib.h>

usage :

void qsort (void *array, size_t count, size_t size, comparison_fn_t compare);

qsort an int array

int compare_int (const void *a, const void *b)

{

    const int *va = (const int *) a;

    const int *vb = (const int *) b;

    return *va-*vb;

}

 

qsort a double array

int compare_double (const void *a, const void *b)

{

    const double *da = (const double *) a;

    const double *db = (const double *) b;

    return (*da > *db) - (*da < *db);

}

Input

Output

Sample Input  Download

Sample Output  Download

Tags

Discuss

Description

A farmer has a crop land. The land consists of stones (S) and holes (H) as illustrated in the following figure. Note that there is only one hole on the land’s first row.

S	S	H	S	S
H	H	H	H	S
H	S	S	S	S
S	H	S	H	H
S	H	S	S	S

 

The farmer decides to water his crops, so he pours some water through the hole on the first row into the land. Assume that the amount of water is infinite, and the water can move in the eight directions: up, down, left, right, upper-left, upper-right, lower-left, lower-right. Please write a program to find out where the water would reach.

For example, you are given a coordinate (0,2) representing the entrance and a table representing the farmer’s land initially:

S	S	H	S	S
H	H	H	H	S
H	S	S	S	S
S	H	S	H	H
S	H	S	S	S

 

After the farmer pours water into the land, water floods throughout the holes of the land. Finally, a portion of the holes will be filled with water like the following table.

S	S	W	S	S
W	W	W	W	S
W	S	S	S	S
S	W	S	H	H
S	W	S	S	S

 

where W means the corresponding hole is filled with water.

Note that

1.      This problem involves three files.

• function.h: Function definition of flooding.
• function.c: Function implementation of flooding.
• 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:

You can use just 4-direction to pass 6 testcases.

function.h

#ifndef FUNCTION_H #define FUNCTION_H void flooding(int R, int C, int pr, int pc); char Map[505][505]; #endif

main.c

Input

The first line has an integer N(1<=N<=100), which means the number of test cases.

For each case, the first line has three integers. The first and second integers R and C (3<=R, C<=500) represent the numbers of rows and columns of the land, respectively. The total number of elements in the land is thus R x C. The third integer X (0<=X<=C-1) means the column of the entrance at the first row. The coordinate of the entrance is thus (0,X). The following R lines, each containing C characters, specify the elements of the farmer’s land.

Output

Print out all elements of the lands row-by-row, and there is a '\n' at the end of each line. The states of all lands should be separated by a new line character (\n).

Sample Input  Download

Sample Output  Download

Partial Judge Code

12104.c

Partial Judge Header

12104.h

Tags

Discuss

Description

Wild cats take care of each other in the wild. However, when winter comes, the preys are not enough to feed all the cats.

In this problem, you are asked to help decide who can enjoy the precious food for three different categories of cats: apprentice, elder, and kitty.

First, the cats dine according to the following order of their occupations:

1. elder

2. kitty

3. apprentice

Then, except that the apprentices have the dining priority of the young over the old, for the other two occupations, the old have higher priority. If the occupations and the ages of two or more cats are the same, they will dine in lexicographic order according to their names.

Input

There are multiple test cases.

The first line of each test case contains two integers N and M, indicating the number of cats and the portions of food respectively, where 0<N,M<=10000.

The next N lines are the information of each cat, including name, occupation, and age.

The length of the names will not exceed 30 letters and will contain no spaces.

Output

Please output the cats that could eat the food in order, each name a line.

Sample Input  Download

Sample Output  Download

Tags

Discuss

Description

Consider a room of size H*W (3<=H < 10000,3<=W < 10000) in which its four walls are covered by mirrors. You need to simulate the laser beam reflection in the room and print out the k-th ( 0 < k < 10000)reflection point.

We assume that the laser beam is always emitted from the left mirror and moves in the upper-right direction at an angle of 45 degrees. More specifically, the starting point can be represented by (r, 1) where 1<r<H. The laser beam will stop if it hits any corner of the room. Also, each point can only reflect the laser beam once. That is, the second time the laser beam hits the same point, the process of reflection stops.
 

For example, if the room size is 5*5 and the laser beam starts at (3,1), then the first point it hits is (3,1), the second point is (1,3), the third point is (3,5), and so on. If k=3 you need to print out (3,5).

If the laser beam stop before the k-th reflection, you need to print out coordinate of the stop point. For example, if k =10 and the first point is (3,1), you need to print out (3,1) because the laser beam stops at (3,1).

Input

The first line is the height and width of the room.

The second line is the starting point (the first reflection point).

The third line is k, which means you need to print out the k-th reflection point.

Output

The coordinate of the k-th reflection point.

Note that you DO NOT need to print a newline character ‘\n’ at the end of the output.

Sample Input  Download

Sample Output  Download

Tags

Discuss

Description

In this problem we simulate a simple text editor. Given two lines of keyboard input texts and commands, output the final text content.

 

The text editing rules are defined as following:

1.

Normal alphabetical input and whitespace input (abcdefg…. and ‘ ‘) in the first line.

And a series of two special commands started with a backslash(/) character in the second line.

2.

The backspace command which deletes a letter before the cursor (/b)

3.

The navigating command which move the cursor to the right(/r)

4.

The cursor is at the beginning of the text initially.

 

For example, for the following input: 

The quick brown fox jumps over the lazy dog
/b/r/r/r/b

You should first store the "The quick brown fox jumps over the lazy dog" in the input char array and set cursor=0.

Then you can simulate the operation of the text editor.

 

The size of the text content is less than or equal to 500 characters, and there will be less than or equal to 500 commands when simulating.

 

Hint:

#include <stdio.h>

#define MAX_SIZE 1001


char content[MAX_SIZE];

char input[MAX_SIZE];

char command[MAX_SIZE];


int main()

{


    gets(input);

    gets(command);


    /* your code here */


    printf("%s", content);


    return 0;

}

Input

The first line is the text content which should be edited.

The second line is the commands.

There is always a valid command(/b /r) right after the backslash character.

Output

The final text content, and there should be no '\n' at the end of the output.

Sample Input  Download

Sample Output  Download

Tags

Discuss

Description

Bonus problem for 2018 Yang Final Exam

There was a lonely singer called Robert who ordered a beer and wanted to write a new song. 

While he writes his song, he often changes the lyrics. 

 

So he asks you to write a program for him.

There are 4 commands:

1. "w" : write the new line in the nth sentence.

2. "r"  : print out all the lyrics in the newest version. 

3. "a" : append the string after the nth sentence (if the nth sentence is blank, then create it.)

4. "s": swap two sentences

 

Hint:

#include <stdio.h> #include <stdlib.h> #include <string.h> #define MAX_X 10000 char *str[MAX_X]; int maxLen[MAX_X]; // str[i]'s max capacity int curLen[MAX_X]; // str[i]'s current length void handlePrint(void) { // TODO // Print out all lyrics in the newest version, ending with a newline. // Skip those lines which don't contain anything. } void handleWrite(char *sentence, int idx) { // TODO // Write string at the sentence on x-th line. (remove the old data in x-th sentence) } void handleAppend(char *sentence, int idx) { // TODO // Append string at the end of the sentence on x-th line. (the sentence might be empty) } void handleSwap(int a, int b) { // TODO // Swap the sentence on a-th line and the sentence on b-th line (they might be empty.) } int main(void) { int n, idx; char inst; char sentence[51]; scanf("%d", &n); // TODO: initialize global variable while (n--) { scanf(" %c", &inst); if (inst == 'r') { handlePrint(); } else if (inst == 'w') { scanf("%d", &idx); getchar(); // consume ' '. scanf("%[^\n]", sentence); handleWrite(sentence, idx); } else if (inst == 'a') { scanf("%d", &idx); getchar(); // consume ' '. scanf("%[^\n]", sentence); handleAppend(sentence, idx); } else if (inst == 's') { int a, b; scanf("%d %d", &a, &b); handleSwap(a, b); } } return 0; }

Input

The first line contains 1 number N, which means the number of instructions. 

Each instruction will be one of the following:

1. w x string: write string at the sentence on x-th line (remove the old data in x-th sentence)
2. a x string: append string at the end of the sentence on x-th line (the sentence might be empty)
3. r: print out all sentences whose length is > 0 (from sentence on first line to sentence on last line, skip those sentence whose length is 0)
4. s a b: swap the sentence on a-th line and the sentence on b-th line (they might be empty.) 

It's guaranteed that

1. 0 <= x < 10000

2. the length of the "string" in each instruction < 50

3. the total length of each sentence might > 50, because of append

4. N <= 1,000,000

Output

If there's a read instruction, output all the setence have so far.

if there's no any words in sentence , do not output anything , even a "\n".

Sample Input  Download

Sample Output  Download

Tags

Discuss