使数字变得神奇的最小操作
给定一个 6 位数字，计算需要替换的最小位数以使该数字具有魔力。如果前三位数字的总和等于后三位数字的总和，则该数字被认为是神奇的。在一次操作中，我们可以在任意位置选择一个数字并将其替换为任意数字。
例子 ：
Input: 123456
Output: 2
Explanation : Replace 4 with 0 and 5 with 0,
then number = 123006, where
1 + 2 + 3 = 0 + 0 + 6,
hence number of replacements
done = 2
Input: 111000
Output: 1
Explanation: Replace 0 with 3, then
number = 111030, where
1 + 1 + 1 = 0 + 3 + 0,
hence number of replacements
done = 1方法 ：
最好的方法是检查所有神奇的数字和所需的替换数量。运行一个生成所有 6 位数字的循环。检查该数字是否神奇，如果是，则只需计算需要完成的替换次数并与 ans 进行比较，如果较小，则将其设为 ans，最后返回 ans。
下面是上述方法的实现。
C++
// CPP program to make a number magical
#include "bits/stdc++.h"
using namespace std;
// function to calculate the minimal changes
int calculate(string s)
{
// maximum digits that can be changed
int ans = 6;
// nested loops to generate all 6
// digit numbers
for (int i = 0; i < 10; ++i) {
for (int j = 0; j < 10; ++j) {
for (int k = 0; k < 10; ++k) {
for (int l = 0; l < 10; ++l) {
for (int m = 0; m < 10; ++m) {
for (int n = 0; n < 10; ++n) {
if (i + j + k == l + m + n) {
// counter to count the number
// of change required
int c = 0;
// if first digit is equal
if (i != s[0] - '0')
c++;
// if 2nd digit is equal
if (j != s[1] - '0')
c++;
// if 3rd digit is equal
if (k != s[2] - '0')
c++;
// if 4th digit is equal
if (l != s[3] - '0')
c++;
// if 5th digit is equal
if (m != s[4] - '0')
c++;
// if 6th digit is equal
if (n != s[5] - '0')
c++;
// checks if less then the
// previous calculate changes
if (c < ans)
ans = c;
}
}
}
}
}
}
}
// returns the answer
return ans;
}
// driver program to test the above function
int main()
{
// number stored in string
string s = "123456";
// prints the minimum operations
cout << calculate(s);
}Java
// java program to make a number magical
import java.io.*;
class GFG {
// function to calculate the minimal changes
static int calculate(String s)
{
// maximum digits that can be changed
int ans = 6;
// nested loops to generate
// all 6 digit numbers
for (int i = 0; i < 10; ++i) {
for (int j = 0; j < 10; ++j) {
for (int k = 0; k < 10; ++k) {
for (int l = 0; l < 10; ++l) {
for (int m = 0; m < 10; ++m) {
for (int n = 0; n < 10; ++n) {
if (i + j + k == l + m + n) {
// counter to count the number
// of change required
int c = 0;
// if first digit is equal
if (i != s.charAt(0) - '0')
c++;
// if 2nd digit is equal
if (j != s.charAt(1) - '0')
c++;
// if 3rd digit is equal
if (k != s.charAt(2) - '0')
c++;
// if 4th digit is equal
if (l != s.charAt(3) - '0')
c++;
// if 5th digit is equal
if (m != s.charAt(4) - '0')
c++;
// if 6th digit is equal
if (n != s.charAt(5) - '0')
c++;
// checks if less then the
// previous calculate changes
if (c < ans)
ans = c;
}
}
}
}
}
}
}
// returns the answer
return ans;
}
// Driver code
static public void main (String[] args)
{
// number stored in string
String s = "123456";
// prints the minimum operations
System.out.println(calculate(s));
}
}
// This code is contributed by vt_m.Python3
# Python 3 program to make a number magical
# function to calculate the minimal changes
def calculate( s):
# maximum digits that can be changed
ans = 6
# nested loops to generate all 6
# digit numbers
for i in range(10):
for j in range(10):
for k in range(10):
for l in range(10):
for m in range(10):
for n in range(10):
if (i + j + k == l + m + n):
# counter to count the number
# of change required
c = 0
# if first digit is equal
if (i != ord(s[0]) - ord('0')):
c+=1
# if 2nd digit is equal
if (j != ord(s[1]) - ord('0')):
c+=1
# if 3rd digit is equal
if (k != ord(s[2]) - ord('0')):
c+=1
# if 4th digit is equal
if (l != ord(s[3]) - ord('0')):
c+=1
# if 5th digit is equal
if (m != ord(s[4]) - ord('0')):
c+=1
# if 6th digit is equal
if (n != ord(s[5]) - ord('0')):
c+=1
# checks if less then the
# previous calculate changes
if (c < ans):
ans = c
# returns the answer
return ans
# driver program to test the above function
if __name__ == "__main__":
# number stored in string
s = "123456"
# prints the minimum operations
print(calculate(s))C#
// C# program to make a number magical
using System;
class GFG {
// function to calculate the minimal changes
static int calculate(string s)
{
// maximum digits that can be changed
int ans = 6;
// nested loops to generate
// all 6 digit numbers
for (int i = 0; i < 10; ++i) {
for (int j = 0; j < 10; ++j) {
for (int k = 0; k < 10; ++k) {
for (int l = 0; l < 10; ++l) {
for (int m = 0; m < 10; ++m) {
for (int n = 0; n < 10; ++n) {
if (i + j + k == l + m + n) {
// counter to count the number
// of change required
int c = 0;
// if first digit is equal
if (i != s[0] - '0')
c++;
// if 2nd digit is equal
if (j != s[1] - '0')
c++;
// if 3rd digit is equal
if (k != s[2] - '0')
c++;
// if 4th digit is equal
if (l != s[3] - '0')
c++;
// if 5th digit is equal
if (m != s[4] - '0')
c++;
// if 6th digit is equal
if (n != s[5] - '0')
c++;
// checks if less then the
// previous calculate changes
if (c < ans)
ans = c;
}
}
}
}
}
}
}
// returns the answer
return ans;
}
// Driver code
static public void Main ()
{
// number stored in string
string s = "123456";
// prints the minimum operations
Console.WriteLine(calculate(s));
}
}
// This code is contributed by vt_m.PHP
Javascript
输出 ：
2时间复杂度： O(10^6)
辅助空间： O(1)