Solution
#include <stdio.h>
const double pi=3.141;
int main ()
{
int r,s;
scanf ("%d%d",&r,&s);
if ((pi*r*r)>(s*s))
{
printf ("I prefer centre 1");
}
else
{
printf ("I prefer centre 2");
}
return 0;
}
Monday, April 29, 2019
Square of a Number
Solution
#include <stdio.h>
int main()
{
int a,squ;
scanf ("%d%d",&a,&squ);
squ=a*a;
printf ("%d",squ);
return 0;
}
Printing Alphabets
Solution
#include<stdio.h>
int main()
{
{
char line[150];
int i, j;
scanf ("%[^\n]s",line);
for(i = 0; line[i] != '\0'; ++i)
{
while (!( (line[i] >= 'a' && line[i] <= 'z') || (line[i] >= 'A' && line[i] <= 'Z') || line[i] == '\0') )
{
for(j = i; line[j] != '\0'; ++j)
{
line[j] = line[j+1];
}
line[j] = '\0';
}
}
puts(line);
return 0;
}
Length of the String
Solution
#include <stdio.h>
#include <string.h>
int main()
{
char a[100];
int length;
scanf ("%[^\n]s",a);
length = strlen(a);
printf("%d\n", length);
return 0;
}
Good Joke!
Vadim and Roman like discussing challenging problems with each other. One day Vadim told his friend following problem:
Given N points on a plane. Each point p is defined by it's two integer coordinates � px and py. The distance between points a and b is min(|ax - bx|, |ay - by|). You should choose a starting point and make a route visiting every point exactly once, i.e. if we write down numbers of points in order you visit them we should obtain a permutation. Of course, overall distance walked should be as small as possible. The number of points may be up to 40.
"40? Maybe 20? Are you kidding?" � asked Roman. "No, it's not a joke" � replied Vadim. So Roman had nothing to do, but try to solve this problem. Since Roman is really weak in problem solving and you are the only friend, except Vadim, with whom Roman can discuss challenging tasks, he has nobody else to ask for help, but you!Input
Input description.
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.The first line of each test case contains a single integer N denoting the number of points on a plane. The following N lines contain two space-separated integers each � coordinates of points.Output
Output description.
Output the answer for every test case in a separate line. The answer for every test case is a permutation of length N. In case there are several solutions that lead to minimal distance walked, you should choose the lexicographically smallest one. Let P denote such permutation. To make output smaller, you should output H(P). H(P) = P1 xor P2 xor ... xor PN. Have a look at the example and it's explanation for better understanding.Constraints
- 1 = T = 10
- 1 = N = 40
- 0 = absolute value of each coordinate = 1000
- 1 = sum over all N in a single test file = 120
Example
Input:
2
2
1 2
0 0
3
3 3
0 0
0 3
Output:
3
0
Solution:
#include<stdio.h>
int main()
{
int T,i,j;
scanf("%d",&T);
int p[40][2];
int N;
int out[T];
for(i=0;i<T;i++)
{
out[i] = 0;
scanf("%d",&N);
for(j=0;j<N;j++)
{
scanf("%d %d",&p[j][0],&p[j][1]);
out[i] = (out[i]^(j+1));
}
}
for(i=0;i<T;i++)
printf("%d\n",out[i]);
return 0;
}
Sunday, April 28, 2019
Sorting
Alice uses the following pseudocode when she needs to sort a permutation of Npositive integers:
procedure Sort(list A) defined as:
list less, greater
if length(A) <= 1 then return A
pivot := A(length(A)+1) / 2
for i := 1 to length(A) do:
Increment(comparison_count)
if Ai < pivot then append Ai to less else if Ai > pivot append Ai to greater
end if
end for
return concatenate(Sort(less), pivot, Sort(greater) )
And now we are interested in the number of comparisons that will be made during the sorting of the given permutation of integers A with the provided code. So, we ask you to find the value of the variable comparison_count after such a sorting.
procedure Sort(list A) defined as:
list less, greater
if length(A) <= 1 then return A
pivot := A(length(A)+1) / 2
for i := 1 to length(A) do:
Increment(comparison_count)
if Ai < pivot then append Ai to less else if Ai > pivot append Ai to greater
end if
end for
return concatenate(Sort(less), pivot, Sort(greater) )
And now we are interested in the number of comparisons that will be made during the sorting of the given permutation of integers A with the provided code. So, we ask you to find the value of the variable comparison_count after such a sorting.
Input
The first line of input consists of a single integer N. The second line of input contains a permutation of N numbers.
Output
Output the number of comparisons on the first and only line of the output.
Example
Input:
5
4 3 5 1 2
Output:
11
Solution
#include <stdio.h>
#include <stdlib.h>
#include <memory.h>
#include <time.h>
#include <limits.h>
int comparison_count;
void debug_array(int *array, int array_size) {
printf("\n");
printf("[");
int i;
for (i = 0; i < array_size; i++) {
printf("%3d", array[i]);
}
printf("]");
printf("\n");
}
unsigned long long bigrand() {
unsigned long long big_rand = 0;
big_rand = rand();
big_rand = big_rand << 32;
big_rand |= rand();
return big_rand;
}
int randint(int start_range, int end_range) {
return (bigrand() % ((end_range + 1) - start_range)) + start_range;
}
void concatenate(int *concatenated,
int concatenated_size,
int *less,
int less_size,
int pivot,
int *greater,
int greater_size) {
int concatenated_idx = 0;
int i;
for (i = 0; i < less_size; i++) {
concatenated[concatenated_idx++] = less[i];
}
concatenated[concatenated_idx++] = pivot;
for (i = 0; i < greater_size; i++) {
concatenated[concatenated_idx++] = greater[i];
}
}
int *sort(int *array, int array_size) {
if (array_size <= 1) {
return array;
}
int pivot = array[(array_size - 1) / 2];
int *less = calloc(array_size, sizeof(int));
int less_idx = 0;
int *greater = calloc(array_size, sizeof(int));
int greater_idx = 0;
memset(less, -1, array_size * sizeof(int));
memset(greater, -1, array_size * sizeof(int));
int i;
for (i = 0; i < array_size; i++) {
comparison_count++;
if(array[i] < pivot) {
less[less_idx++] = array[i];
} else if (array[i] > pivot) {
greater[greater_idx++] = array[i];
}
}
int concatenated_size = less_idx + greater_idx + 1;
int *concatenated = calloc(concatenated_size, sizeof(int));
concatenate(concatenated,
concatenated_size,
sort(less, less_idx),
less_idx,
pivot,
sort(greater, greater_idx),
greater_idx);
return concatenated;
}
float calc_elapsed_time(clock_t time_s, clock_t time_e) {
return (((float)time_e - (float)time_s) / CLOCKS_PER_SEC) * 1000;
}
int main(int argc, const char * argv[])
{
int N;
scanf("%d", &N);
int *A = calloc(N, sizeof(int));
int i;
for (i = 0; i < N; i++) {
scanf("%d", &A[i]);
}
comparison_count = 0;
int *sorted = sort(A, N);
free(sorted);
printf("%d\n", comparison_count);
return 0;
}
Chef And his Cake
Chef�s girlfriend's birthday is near, so he wants to surprise her by making a special cake for her. Chef knows that his girlfriend likes cherries on the cake, so he puts cherries on the top of the cake, but he was not satisfied. Therefore, he decided to replace some of the cherries to make a beautiful pattern. However, Chef has a lot of other work to do so he decided to ask for your help.
The cherries are of two colors red and green. Now Chef wants the cherries to be placed in such a way that each cherry of one color must be adjacent to only cherries of the other color, two cherries are adjacent if they share a side. Now Chef has asked for your help in making that pattern on the cake.
You can replace any cherry of given color with the other color. But there is a cost for each replacement: if you replace a red cherry with a green one, the cost is 5units and if you replace a green cherry with a red one, the cost is 3 units.
Help your friend Chef by making the cake special with minimum cost.
Input
- The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
- The first line of each test case contains two space-separated integers N and M, where N � M are the dimensions of the cake.
- Each of the next N lines contains a string of length M.
Output
For each test case, output the minimum cost required to make the cake special.
Constraints
- 1 = T = 100
- 1 = N, M = 100
- each string consists only of letters 'R' and 'G' denoting red and green cherries respectively
Example
Input:8
2
4 5
RGRGR
GRGRG
RGRGR
GRGRG
2 3
RRG
GGR
Output:
0
Solution
#include <stdio.h>
int main(void) {
// your code goes here
int T;
scanf(" %d",&T);
while(T--){
int x,y,cost1=0,cost2=0;
scanf("%d %d",&x,&y);
char cake[x][y];
for(int i=0; i<x; i++)
scanf(" %s",cake[i]);
for(int i=0;i<x;i++)
{
for(int j=0;j<y;j++)
{
if((i+j)%2==0 && cake[i][j]=='G')
cost1 +=3;
else if((i+j)%2==0 && cake[i][j]=='R')
cost2 +=5;
else if((i+j)%2==1 && cake[i][j]=='R')
cost1 +=5;
else if((i+j)%2==1 && cake[i][j]=='G')
cost2 +=3;
}
}
(cost1 < cost2)? printf("%d\n",cost1) : printf("%d\n",cost2);
}
return 0;
}
Trip
You are starting out on a long (really long) trip. On the way, there are N gas stations, the locations of which are given as a1,a2,�,aN. Initially you are located at the gas station at a1, and your destination is at location a_N. Your car can only store enough fuel to travel atmost M units without refilling. You can stop at any station and refill the car by any amount. Now you wish to plan your trip such that the number of intermediate stops needed to reach the destination is minimum, and also how many ways are there to plan your trip accordingly.
Input :
The first line two space seperated integers N and M. N lines follow, and the ith line has the value ai (0 <= ai <= 1000000000). The input will be such that a solution will always exist.
Output :
Output two space seperated integers : The least number of stops, and the number of ways to plan the trip which uses the least number of stops. Output this value modulo 1000000007.
Sample Input :
6 3
0
1
3
4
7
10
Sample Output :
3 2
Constraints :
2 <= N <= 1000000
1 <= M <= 1000
a_1 < a_2 < .. < a_N
Solution
#include<stdio.h>
#include<stdlib.h>
#define MOD 1000000007
#define BUF 4096
int arr[1000000];
int temp1[1000000];
int temp2[1000000];
char ibuf[BUF];
int ipt = BUF;
int readInt()
{
while (ipt < BUF && ibuf[ipt] < '0')
ipt++;
if (ipt == BUF)
{
fread(ibuf, 1, BUF, stdin);
ipt = 0;
while (ipt < BUF && ibuf[ipt] < '0')
ipt++;
}
int n = 0;
while (ipt < BUF && ibuf[ipt] >= '0')
n = (n*10)+(ibuf[ipt++]-'0');
if (ipt == BUF)
{
fread(ibuf, 1, BUF, stdin);
ipt = 0;
while (ipt < BUF && ibuf[ipt] >= '0')
n = (n*10)+(ibuf[ipt++]-'0');
}
return n;
}
int main()
{
int m,n,i,j,min,k;
long long count=0;
n=readInt();
m=readInt();
for(i=0;i<n;i++)
arr[i]=readInt();
temp2[0]=1;
j=k=count=0;
for(i=1;i<n;i++)
{
while(arr[i]-arr[j] > m)
count-=temp2[j++];
temp1[i]=temp1[j]+1;
while(temp1[k]==temp1[j])
count+=temp2[k++];
temp2[i]=count%MOD;
}
printf("%d %d\n",temp1[n-1]-1,temp2[n-1]);
return 0;
}Saturday, April 27, 2019
Occurrence
Solutions
#include <stdio.h>
#define MAX 100
int main()
{
int arr[MAX],n,i;
int num,count;
scanf("%d",&n);
for(i=0;i< n;i++)
{
scanf("%d",&arr[i]);
}
scanf("%d",&num);
count=0;
for(i=0;i< n;i++)
{
if(arr[i]==num)
count++;
}
printf("%d",count);
return 0;
}
Pogo Stick Jump
Raju lives in a colony. On his 9th birthday, his father gift him a Pogo Stick. He
is so excited to play with pogo stick. The pogo stick
moves one unit per jump. He wanders around his house jumping with pogo sticks. He
wants to show the pogo stick to his friends and
decide to go using pogo sticks. Write a program to find number of jumps needed to
reach his friends house. Assume that Rajus house is
in the location (3,4). Input and Output Format: Input consists of two integers x,
y. The x and y corresponds to x and y coordinates of his
friends house. Output is an integer - the number of jumps he needs to reach his
friends house.
Solution
#include <stdio.h>
#include <math.h>
int main ()
{
int x1=3,y1=4,x2,y2;
float power,power1,power2,sqr;
scanf ("%d",&x2);
scanf ("%d",&y2);
power1=pow((x2-x1),2);
power2=pow((y2-y1),2);
power=power1+power2;
sqr=sqrt(power);
printf ("Raju needs %0.f jumps",sqr);
return 0;
}
Sample Input
5 6
Sample Output
Raju needs 3 jumps
is so excited to play with pogo stick. The pogo stick
moves one unit per jump. He wanders around his house jumping with pogo sticks. He
wants to show the pogo stick to his friends and
decide to go using pogo sticks. Write a program to find number of jumps needed to
reach his friends house. Assume that Rajus house is
in the location (3,4). Input and Output Format: Input consists of two integers x,
y. The x and y corresponds to x and y coordinates of his
friends house. Output is an integer - the number of jumps he needs to reach his
friends house.
Solution
#include <stdio.h>
#include <math.h>
int main ()
{
int x1=3,y1=4,x2,y2;
float power,power1,power2,sqr;
scanf ("%d",&x2);
scanf ("%d",&y2);
power1=pow((x2-x1),2);
power2=pow((y2-y1),2);
power=power1+power2;
sqr=sqrt(power);
printf ("Raju needs %0.f jumps",sqr);
return 0;
}
Sample Input
5 6
Sample Output
Raju needs 3 jumps
Take a count
Sheela gave a number to Jonnes and ask him to count the number of digits in a number. Sheela tries one simple logic to do that. So she decided to apply this concept in C.
Test Case 1
Input (stdin)
3452
Expected Output
4
Test Case 2
Input (stdin)
562356
Expected Output
6
Solution
#include <stdio.h>
int main()
{
long long num;
int count = 0;
scanf("%lld", &num);
while(num != 0)
{
count++;
num /= 10;
}
printf("%d", count);
return 0;
}
Test Case 1
Input (stdin)
3452
Expected Output
4
Test Case 2
Input (stdin)
562356
Expected Output
6
Solution
#include <stdio.h>
int main()
{
long long num;
int count = 0;
scanf("%lld", &num);
while(num != 0)
{
count++;
num /= 10;
}
printf("%d", count);
return 0;
}
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