UVA10880 Colin and Ryan整除

Posted 2021-06-14 海岛Blog

“As a special bonus, if they order
by midnight, what would they receive?
Colin: They’d receive it earlier than if
they’d ordered it later.
Songs of the Butcher

    Colin and Ryan had a party. They baked C cookies and invited G guests. Each guest ate Q cookies, and R cookies were left (R < Q).
Input
The first line of input gives the number of cases, N. N test cases follow. Each one is a line containing C and R (at most 2000000000).
Output
For each test case, output one line containing ‘Case #x:’ followed by Q — the number of cookies each guest ate. If there are multiple answers, print them in increasing order, separated by spaces. Do not print trailing spaces. Print a ‘0’ in the case when R = C.
Sample Input
4
10 0
13 2
300 98
1000 997
Sample Output
Case #1: 1 2 5 10
Case #2: 11
Case #3: 101 202
Case #4:

问题链接：UVA10880 Colin and Ryan
问题简述：总共C块饼干，请G个客人吃，每人吃Q块，余下R块。给定C和R，问每位客人可以吃多少块饼干，有多种解，从小到大输出这些解。
问题分析：整除问题，求C-R的因子即可，用STL的set来排序。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA10880 Colin and Ryan */

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int t, caseno = 0;
    scanf("%d", &t);
    while (t--) {
        int c, r;
        scanf("%d%d", &c, &r);

        int diff = c - r;

        printf("Case #%d:", ++caseno);
        if (diff == 0)
            printf(" 0\\n");
        else {
            set<int> ans;
            for (int i = 1; i * i <= diff; i++)
                if (diff % i == 0) {
                    ans.insert(diff / i);
                    ans.insert(i);
                }
            for (set<int>::iterator iter = ans.begin(); iter != ans.end(); iter++)
                if (*iter > r) printf(" %d", *iter);
            printf("\\n");
        }
    }

    return 0;
}