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  • PAT Advanced 1069 The Black Hole of Numbers (20分)

    For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in non-increasing order first, and then in non-decreasing order, a new number can be obtained by taking the second number from the first one. Repeat in this manner we will soon end up at the number 6174 -- the black hole of 4-digit numbers. This number is named Kaprekar Constant.

    For example, start from 6767, we'll get:

    7766 - 6677 = 1089
    9810 - 0189 = 9621
    9621 - 1269 = 8352
    8532 - 2358 = 6174
    7641 - 1467 = 6174
    ... ...
    
     

    Given any 4-digit number, you are supposed to illustrate the way it gets into the black hole.

    Input Specification:

    Each input file contains one test case which gives a positive integer N in the range (.

    Output Specification:

    If all the 4 digits of N are the same, print in one line the equation N - N = 0000. Else print each step of calculation in a line until 6174 comes out as the difference. All the numbers must be printed as 4-digit numbers.

    Sample Input 1:

    6767
    
     

    Sample Output 1:

    7766 - 6677 = 1089
    9810 - 0189 = 9621
    9621 - 1269 = 8352
    8532 - 2358 = 6174
    
     

    Sample Input 2:

    2222
    
     

    Sample Output 2:

    2222 - 2222 = 0000

    这道题主要考察了字符串的转换,我们使用转换字符串的方式,可以快速得到答案,

    使用do while循环,避免了0或刚开始是6767不输出情况

    #include <iostream>
    #include <algorithm>
    using namespace std;
    int main() {
        int N;
        scanf("%d", &N);
        string str, rev;
        do {
            str = to_string(N);
            while(str.length() != 4) str = "0" + str;
            sort(str.begin(), str.end());
            rev = str;
            reverse(rev.begin(), rev.end());
            N = stoi(rev) - stoi(str);
            printf("%s - %s = %04d
    ", rev.c_str(), str.c_str(), N);
        }while(N != 0 && N != 6174);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/littlepage/p/12251833.html
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