UVA11371 Number Theory for Newbies水题

Posted 2021-06-29 海岛Blog

Given any positive integer, if we permute its digits, the difference between the number we get and the given number will always be divisible by 9.
For example, if the given number is 123, we may rearrange the digits to get 321. The difference = 321 - 123 = 198, which is a multiple of 9 (198 = 9 × 22).
We can prove this fact fairly easily, but since we are not having a maths contest, we instead try to illustrate this fact with the help of a computer program.
Input
Each line of input gives a positive integer n (≤ 2000000000). You are to find two integers a and b formed by rearranging the digits of n, such that a − b is maximum. a and b should NOT have leading zeros.
Output
You should then show that a − b is a multiple of 9, by expressing it as ‘9 * k’, where k is an integer. See the sample output for the correct output format.
Sample Input
123
2468
Sample Output
321 - 123 = 198 = 9 * 22
8642 - 2468 = 6174 = 9 * 686

问题链接：UVA11371 Number Theory for Newbies
问题简述：给定一个正整数n，将其每位数字重新组合成，可以得到整数a和b，使得a-b最大且满足是9的倍数。
问题分析：简单题，不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA11371 Number Theory for Newbies */

#include <bits/stdc++.h>

using namespace std;

char s[10 + 1];

int main()
{
    while (scanf("%s", s) == 1) {
        int len = strlen(s);
        sort(s, s + len);
        for (int i = 0; i < len; i++)
            if (s[i] != '0') {
                swap(s[0], s[i]);
                break;
            }

        long long a, b;
        sscanf(s, "%lld", &b);
        sort(s, s + len);
        for (int i = 0, j = len - 1; i < j; i++, j--)
            swap(s[i], s[j]);
        sscanf(s, "%lld", &a);

        printf("%lld - %lld = %lld = 9 * %lld\\n", a, b, a - b, (a - b) / 9);
    }

    return 0;
}