1 条题解

  • -2

    #include<iostream>
    #include<cmath>
    using namespace std;

    int chongzu_max(int n);
    int chongzu_min(int n);

    int main()
    {
    int n;
    cin>>n;
    int n2 = n;
    while(n2 != 495)
    {
    int n0 = chongzu_max(n2);
    int n1 = chongzu_min(n2);
    n2 = n0 - n1;
    cout<<n0<<" - "<<n1<<" = "<<n2<<endl;
    if(n2 < 100 && n2 >= 10) n2 *= 10;
    if(n2 < 10) n *= 100;
    if(n2 == 0) return 0;
    }
    return 0;
    }

    int chongzu_max(int x)
    {
    int a = x % 10;
    x /= 10;
    int b = x % 10;
    x /= 10;
    int c = x % 10;
    int n[6];
    n[0] = a * 100 + b * 10 + c;
    n[1] = a * 100 + c * 10 + b;
    n[2] = b * 100 + a * 10 + c;
    n[3] = b * 100 + c * 10 + a;
    n[4] = c * 100 + a * 10 + b;
    n[5] = c * 100 + b * 10 + a;
    int max = 0;
    for(int i = 0; i <= 5; i++)
    if(n[i] > max) max = n[i];
    return max;
    }

    int chongzu_min(int x)
    {
    int a = x % 10;
    x /= 10;
    int b = x % 10;
    x /= 10;
    int c = x % 10;
    int n[6];
    n[0] = a * 100 + b * 10 + c;
    n[1] = a * 100 + c * 10 + b;
    n[2] = b * 100 + a * 10 + c;
    n[3] = b * 100 + c * 10 + a;
    n[4] = c * 100 + a * 10 + b;
    n[5] = c * 100 + b * 10 + a;
    int min = n[0];
    for(int i = 1; i <= 5; i++)
    if(n[i] < min) min = n[i];
    return min;
    }

  • 1

信息

ID
1204
难度
1
分类
(无)
标签
递交数
31
已通过
26
通过率
84%
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