给定一个包含字符串的大小为N的数组arr[] ，任务是计算字符的 ASCII 值之和等于阿姆斯特朗数或质数的字符串的数量。
例子：
Input: arr[] = {“hello”, “nace”}
Output:
Number of Armstrong Strings are: 1
Number of Prime Strings are: 0
Explanation: Sum of ASCII values of characters of each string is: {532, 407}, out of which 407 is an Armstrong Number, and none of them is a Prime Number.
Hence, the armstrong valued string is “nace”.
Input: arr[] = {“geeksforgeeks”, “a”, “computer”, “science”, “portal”, “for”, “geeks”}
Output:
Number of Armstrong Strings are: 0
Number of Prime Strings are: 2
Explanation: Sum of ASCII values of characters of each string is: {1381, 97, 879, 730, 658, 327, 527}, out of which 1381 and 97 are Prime Numbers, and none of them is an Armstrong Number.
Hence, prime valued strings are “geeksforgeeks” and “a”.
方法：这个问题可以通过计算每个字符串的ASCII值来解决。请按照以下步骤解决此问题：
- 将两个变量countPrime和countArmstrong初始化为0 ，以存储 Prime 和 Armstrong 值字符串的计数。
- 使用变量i迭代索引范围[0, N – 1]并执行以下步骤：
- 将当前字符串arr[i]的字符的 ASCII 值总和存储在一个变量中，比如val 。
- 如果数字val是阿姆斯壮数，则将 countArmstrong增加1 。
- 如果数字val是质数，则将 countPrime增加1 。
- 打印countPrime和countArmstrong的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if a
// number is prime number
bool isPrime(int num)
{
// Define a flag variable
bool flag = false;
if (num > 1)
{
// Check for factors of num
for(int i = 2; i < num; i++)
{
// If factor is found,
// set flag to True and
// break out of loop
if ((num % i) == 0)
{
flag = true;
break;
}
}
}
// Check if flag is True
if (flag)
return false;
else
return true;
}
// Function to calculate
// order of the number x
int order(int x)
{
int n = 0;
while (x != 0)
{
n = n + 1;
x = x / 10;
}
return n;
}
// Function to check whether the given
// number is Armstrong number or not
bool isArmstrong(int x)
{
int n = order(x);
int temp = x;
int sum1 = 0;
while (temp != 0)
{
int r = temp % 10;
sum1 = sum1 + pow(r, n);
temp = temp / 10;
}
// If the condition satisfies
return (sum1 == x);
}
// Function to count
// Armstrong valued strings
int count_armstrong(vector li)
{
// Stores the count of
// Armstrong valued strings
int c = 0;
// Iterate over the list
for(string ele : li)
{
// Store the value
// of the string
int val = 0;
// Find value of the string
for(char che:ele)
val += che;
// Check if it an Armstrong number
if (isArmstrong(val))
c += 1;
}
return c;
}
// Function to count
// prime valued strings
int count_prime(vector li)
{
// Store the count of
// prime valued strings
int c = 0;
// Iterate over the list
for(string ele:li)
{
// Store the value
// of the string
int val = 0;
// Find value of the string
for(char che : ele)
val += che;
// Check if it
// is a Prime Number
if (isPrime(val))
c += 1;
}
return c;
}
// Driver code
int main()
{
vector arr = { "geeksforgeeks", "a", "computer",
"science", "portal", "for", "geeks"};
// Function Call
cout << "Number of Armstrong Strings are: "
<< count_armstrong(arr) << endl;
cout << "Number of Prime Strings are: "
<< count_prime(arr) << endl;
}
// This code is contributed by mohit kumar 29 Python3
# Python program for the above approach
# Function to check if a
# number is prime number
def isPrime(num):
# Define a flag variable
flag = False
if num > 1:
# Check for factors of num
for i in range(2, num):
# If factor is found,
# set flag to True and
# break out of loop
if (num % i) == 0:
flag = True
break
# Check if flag is True
if flag:
return False
else:
return True
# Function to calculate
# order of the number x
def order(x):
n = 0
while (x != 0):
n = n + 1
x = x // 10
return n
# Function to check whether the given
# number is Armstrong number or not
def isArmstrong(x):
n = order(x)
temp = x
sum1 = 0
while (temp != 0):
r = temp % 10
sum1 = sum1 + r**n
temp = temp // 10
# If the condition satisfies
return (sum1 == x)
# Function to count
# Armstrong valued strings
def count_armstrong(li):
# Stores the count of
# Armstrong valued strings
c = 0
# Iterate over the list
for ele in li:
# Store the value
# of the string
val = 0
# Find value of the string
for che in ele:
val += ord(che)
# Check if it an Armstrong number
if isArmstrong(val):
c += 1
return c
# Function to count
# prime valued strings
def count_prime(li):
# Store the count of
# prime valued strings
c = 0
# Iterate over the list
for ele in li:
# Store the value
# of the string
val = 0
# Find value of the string
for che in ele:
val += ord(che)
# Check if it
# is a Prime Number
if isPrime(val):
c += 1
return c
# Driver code
arr = ["geeksforgeeks", "a", "computer",
"science", "portal", "for", "geeks"]
# Function Call
print("Number of Armstrong Strings are:", count_armstrong(arr))
print("Number of Prime Strings are:", count_prime(arr))时间复杂度： O(N*M)，其中 M 是数组arr[] 中最长字符串的长度
辅助空间： O(1)