Answer:
#include <bits/stdc++.h>
using namespace std;
bool isPalindrome(string str)
{
char a,b;
int length = str.length();
for (int i = 0; i < length / 2; i++)
{
a=tolower(str[i]);//Converting both first characters to lowercase..
b=tolower(str[length-1-i]);
if (b != a )
return false;
}
return true;
}
int main() {
string t1;
cin>>t1;
if(isPalindrome(t1))
cout<<"The string is Palindrome"<<endl;
else
cout<<"The string is not Palindrome"<<endl;
return 0;
}
Output:-
Enter the string
madam
The string is Palindrome
Enter the string
abba
The string is Palindrome
Enter the string
22
The string is Palindrome
Enter the string
67876
The string is Palindrome
Enter the string
444244
The string is not Palindrome
Explanation:
To ignore the cases of uppercase and lower case i have converted every character to lowercase then checking each character.You can convert to uppercase also that will also work.
Answer:
age = int(input("Enter your age: "))
charge = 0
if age > 55:
charge = 10
if 21 <= age <= 54:
charge = 15
if 13 <= age <= 20:
charge = 10
if 3 <= age <= 12:
charge = 5
if 0 <= age < 3:
charge = 0
print(charge)
Explanation:
*It is in Python.
Ask the user for the age
Check the each given range and set the charge accordingly using if statements
Print the charge
Answer:
2^7= 128
Explanation:
An instruction format characterizes the diverse part of a guidance. The fundamental segments of an instruction are opcode and operands. Here are the various terms identified with guidance design: Instruction set size tells the absolute number of guidelines characterized in the processor. Opcode size is the quantity of bits involved by the opcode which is determined by taking log of guidance set size. Operand size is the quantity of bits involved by the operand. Guidance size is determined as total of bits involved by opcode and operands.
Answer:
Check the explanation
Explanation:
Algorithm for solving flood condition:
We suggest an algorithm to resolve the flood condition by creating a flow network graph.
Let us assume for every patient "p" there is a node "2" and for every hospital "h" there is a node "uh" and there is an edge ()T, uh) exist between patient "p" and hospital "h" with flow capacity of 1 iff patient "p" is reachable to hospital "h" within a half-hour.
Then source node "s" is made between all the patient-nodes by an edge with flow capacity of 1 and then the sink "t" is made by linking all the hospital nodes by an edge with capacity "[n/k]".
There is an approach to send patients to hospitals: when there is a source "s" to sink "t" flow of "n". We can send 1 flow-unit from source "s" to sink "t" along the paths (s, yp, uh, t) whenever a probable approach is available to send patients.
This approach of sending patients to hospitals doesn't break the capacity limitation of edges. Hence we can send patient "p" to hospital "h" with 1 flow- unit if edge(m uh) permits at least 1 flow- unit.
The running-time of this algorithm is found by finding the time needed to solve max-flow graph with nodes O(n+k) and edges O(
) edges.