计算 ASCII 值为质数的字符串中的字符数
原文:https://www . geesforgeks . org/count-字符串中的字符-其 ascii 值为-prime/
给定一根绳子 S 。任务是计算并打印 ASCII 值为质数的字符串中的字符数。
示例:
输入: S = "geeksforgeeks" 输出:3 “g”、“e”和“k”是 ASCII 值为质数的唯一字符，即分别为 103、101 和 107。
输入:S = " abcdefghijklmnopqrstuvwxyz " T3】输出: 6
方法:思路是使用厄拉多塞的筛生成字符串 S 字符最大 ASCII 值以内的所有素数。现在，迭代字符串并获取每个字符的 ASCII 值。如果 ASCII 值是质数，则增加计数。最后，打印计数。
下面是上述方法的实现:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
#define max_val 257
// Function to find prime characters in the string
int PrimeCharacters(string s)
{
// USE SIEVE TO FIND ALL PRIME NUMBERS LESS
// THAN OR EQUAL TO max_val
// Create a Boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
vector<bool> prime(max_val + 1, true);
// 0 and 1 are not primes
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p)
prime[i] = false;
}
}
int count = 0;
// Traverse all the characters
for (int i = 0; i < s.length(); ++i) {
if (prime[int(s[i])])
count++;
}
return count;
}
// Driver program
int main()
{
string S = "geeksforgeeks";
// print required answer
cout << PrimeCharacters(S);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of above approach
class Solution
{
static final int max_val=257;
// Function to find prime characters in the String
static int PrimeCharacters(String s)
{
// USE SIEVE TO FIND ALL PRIME NUMBERS LESS
// THAN OR EQUAL TO max_val
// Create a Boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
boolean prime[]= new boolean[max_val+1];
//initialize the value
for(int i=0;i<=max_val;i++)
prime[i]=true;
// 0 and 1 are not primes
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p)
prime[i] = false;
}
}
int count = 0;
// Traverse all the characters
for (int i = 0; i < s.length(); ++i) {
if (prime[(int)(s.charAt(i))])
count++;
}
return count;
}
// Driver program
public static void main(String args[])
{
String S = "geeksforgeeks";
// print required answer
System.out.print( PrimeCharacters(S));
}
}
//contributed by Arnab Kundu
Python 3
# Python3 implementation of above approach
from math import sqrt
max_val = 257
# Function to find prime characters in the string
def PrimeCharacters(s) :
# USE SIEVE TO FIND ALL PRIME NUMBERS LESS
# THAN OR EQUAL TO max_val
# Create a Boolean array "prime[0..n]". A
# value in prime[i] will finally be false
# if i is Not a prime, else true.
prime = [True] * (max_val + 1)
# 0 and 1 are not primes
prime[0] = False
prime[1] = False
for p in range(2, int(sqrt(max_val)) + 1) :
# If prime[p] is not changed, then
# it is a prime
if (prime[p] == True) :
# Update all multiples of p
for i in range(2*p ,max_val + 1, p) :
prime[i] = False
count = 0
# Traverse all the characters
for i in range(len(s)) :
if (prime[ord(s[i])]) :
count += 1
return count
# Driver program
if __name__ == "__main__" :
S = "geeksforgeeks";
# print required answer
print(PrimeCharacters(S))
# This code is contributed by Ryuga
C
// C# implementation of above approach
using System;
class GFG{
static readonly int max_val = 257;
// Function to find prime characters in the String
static int PrimeCharacters(String s)
{
// USE SIEVE TO FIND ALL PRIME NUMBERS LESS
// THAN OR EQUAL TO max_val
// Create a Boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
bool []prime = new bool[max_val + 1];
//initialize the value
for(int i = 0; i <= max_val; i++)
prime[i] = true;
// 0 and 1 are not primes
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= max_val; p++)
{
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p)
prime[i] = false;
}
}
int count = 0;
// Traverse all the characters
for (int i = 0; i < s.Length; ++i)
{
if (prime[(int)(s[i])])
count++;
}
return count;
}
// Driver Code
public static void Main()
{
String S = "geeksforgeeks";
// print required answer
Console.Write( PrimeCharacters(S));
}
}
// This code is contributed by PrinciRaj1992
java 描述语言
<script>
// JavaScript implementation of above approach
const max_val = 257;
// Function to find prime characters in the String
function PrimeCharacters(s) {
// USE SIEVE TO FIND ALL PRIME NUMBERS LESS
// THAN OR EQUAL TO max_val
// Create a Boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
var prime = new Array(max_val + 1);
//initialize the value
for (var i = 0; i <= max_val; i++) prime[i] = true;
// 0 and 1 are not primes
prime[0] = false;
prime[1] = false;
for (var p = 2; p * p <= max_val; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] === true) {
// Update all multiples of p
for (var i = p * 2; i <= max_val; i += p) prime[i] = false;
}
}
var count = 0;
// Traverse all the characters
for (var i = 0; i < s.length; ++i) {
if (prime[s[i].charCodeAt(0)]) count++;
}
return count;
}
// Driver Code
var S = "geeksforgeeks";
// print required answer
document.write(PrimeCharacters(S));
// This code is contributed by rdtank.
</script>
Output:
8
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