#include <stdio.h> int paixu(int num); int paixiao(int num); int main() { int n; scanf("%d",&n); while(n--) { int cum=0; int num,a,b; scanf("%d",&num); while(1) { b=paixu(num); a=paixiao(num); if(a-b==num) { cum++; printf("%d\n",cum); break; } num=a-b; cum++; }

} return 0;

} int paixu(int num) { int a[10]; int cont=0,total=1; while(num>0) { a[cont++]=num%10; num/=10; } for(int i=0;i<cont;i++) { for(int j=0;j<cont-1;j++) { int temp; if(a[j]>a[j+1]) { temp=a[j]; a[j]=a[j+1]; a[j+1]=temp;//从小到大排序 } } } for(int i=0;i<cont;i++) { total=total10+a[i]; } return total; } int paixiao(int num) { int a[10]; int cont=0,total=1; while(num>0) { a[cont++]=num%10; num/=10; } for(int i=0;i<cont;i++) { for(int j=0;j<cont-1;j++) { int temp; if(a[j]<a[j+1]) { temp=a[j]; a[j]=a[j+1]; a[j+1]=temp;//从大到小排序 } } } for(int i=0;i<cont;i++) { total=total10+a[i]; } return total; }

#include<stdio.h> int main() { int n,m,a,b,oi[4]; int i,j,count; scanf("%d", &n);

while(n--) { scanf("%d", &m); count=0; while (m!= 6174) { oi[0] = m / 1000; oi[1] = m / 100 % 10; oi[2] = m / 10 % 10; oi[3] = m % 10; for (i = 0; i < 3; i++) { for (j = 0; j < 3 - i; j++) { if (oi[j] >oi[j + 1]) { int temp; temp = oi[j]; oi[j] = oi[j + 1]; oi[j + 1] = temp; } } } a = oi[3] * 1000 + oi[2] * 100 + oi[1] * 10 + oi[0]; b = oi[0] * 1000 + oi[1] * 100 + oi[2] * 10 + oi[3]; m=a-b; count++; } printf("%d\n", count+1); } return 0;

}

//经典的数字黑洞问题
#include<stdio.h>
int main()
{
    int c;
    scanf("%d",&c);
    while(c--)
    {
        int a[4];
        int i,j,m,n,t,s,count=0;
        scanf("%d",&s);
        while(s!=6174)
        {
            a[0]=s/1000;
            a[1]=s/100%10;
            a[2]=s/10%10;
            a[3]=s%10;
            for(i=0;i<3;i++)
            {
                for(j=i+1;j<4;j++)
                {
                    if(a[j]>a[i])
                    {
                        t=a[j];
                        a[j]=a[i];
                        a[i]=t;
                    }
                }
            }
             m=1000*a[0]+100*a[1]+10*a[2]+a[3];
             n=1000*a[3]+100*a[2]+10*a[1]+a[0];
             s=m-n;
             count++;
        }
        printf("%d\n",count+1);
    }
    return 0;
}

避免数位排序的方法：桶排。

#include <cstdio>
#include <cstring>

int digit[10];

int process(int m){
    memset(digit, 0, sizeof(digit));
    ++ digit[m / 1000];
    ++ digit[(m % 1000) / 100];
    ++ digit[(m % 100) / 10];
    ++ digit[m % 10];
    int big = 0, small = 0, timer = 1;
    for(int i = 0; i <= 9; ++ i){
        while(digit[i]){
            big += i * timer;
            small += i * 1000 / timer;
            timer *= 10;
            -- digit[i];
        }
    }
    return big - small;
}

int main(){
    int n, i;
    int num, cnt = 0;
    scanf("%d", &n);
    for(i = 1; i <= n; ++ i){
        scanf("%d", &num);
        while(num != 6174){
            num = process(num);
            ++ cnt;
        }
        printf("%d\n", cnt + 1);
        cnt = 0;
    }
}