question
For any four-digit positive integer (except those with the same four digits), if the digits are combined into a maximum and minimum number, subtract the two numbers and repeat this process, we will get 6174.
1. Analyze the problem
- Known: four-digit positive integer
- Unknown: The process of getting 6174
- Relationship: Each digit forms a maximum number and a minimum number, and the two numbers are subtracted.
2. Define variables
//二、数据定义
int n;
3. Enter data
//三、数据输入
cin>>n;
4.Data calculation
Because the number needs to be reassembled, disassembly is definitely used, and short division is preferred.
temp: actually the value of n, used for short division iteration loops.
a[4]: An array is used here to store the number obtained by splitting.
c: Record array subscript.
int temp=n,max,min,result,c=0,a[4]={
0};
while(temp>0){
a[c]=temp%10;
++c;
temp/=10;
}
If I need to form a maximum number max and a minimum number min, then the way I think of is to sort the array, then the maximum number and the minimum number will be known.
for(int i=0;i<3;i++){
for(int j=0;j<3-i;j++){
if(a[j]>a[j+1]){
int t=a[j];
a[j]=a[j+1];
a[j+1]=t;
}
}
}
max=a[3]*1000+a[2]*100+a[1]*10+a[0];
min=a[0]*1000+a[1]*100+a[2]*10+a[3];
Subtract the result of the two numbers and output it.
result=max-min;
cout<<max<<"-"<<min<<"="<<result<<endl;
Repeat this process until this number = 6174.
//四、数据计算
while(6174!=n){
n=result;
}
5. Output results
Complete code.
#include<iostream>
using namespace std;
int main(){
//一、分析问题
//已知：四位正整数
//未知：得到6174的过程
//关系：各位数字组成一个最大数和最小数，两数相减
//二、数据定义
int n;
//三、数据输入
cin>>n;
//四、数据计算
while(6174!=n){
int temp=n,max,min,result,c=0,a[4]={
0};
while(temp>0){
a[c]=temp%10;
++c;
temp/=10;
}
for(int i=0;i<3;i++){
for(int j=0;j<3-i;j++){
if(a[j]>a[j+1]){
int t=a[j];
a[j]=a[j+1];
a[j+1]=t;
}
}
}
max=a[3]*1000+a[2]*100+a[1]*10+a[0];
min=a[0]*1000+a[1]*100+a[2]*10+a[3];
result=max-min;
cout<<max<<"-"<<min<<"="<<result<<endl;
n=result;
}
//五、输出结果
return 0;
}