Satisfactory Pairs

最新推荐文章于 2022-06-11 14:21:08 发布

原创最新推荐文章于 2022-06-11 14:21:08 发布 · 330 阅读

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本文介绍了一种算法，该算法用于找出所有符合条件的正整数对(a, b)，使得方程x*a + y*b = n至少有一个正整数解(x, y)。通过预处理得到n以内所有数的因数，并使用暴力枚举的方法来寻找满足条件的整数对。

Given a positive integer n, find and print the number of pairs of positive integers(a,b), where(a < b) , that exist such that the equation x * a + y * b = n (where x and yare positive integers) has at least one solution.

Input Format

A single positive integer denoting n.

Constraints

4 <= n <= 3 * 10 ^ 5;

Output Format

Print a single integer denoting the number of such pairs.

Sample Input 0

Sample Output 0

Explanation 0

There are two such (a, b) pairs: (1,2) and (1, 3).

预处理N以内所有数的约数，暴力枚举。

#include <bits/stdc++.h>

using namespace std;
vector<int>::iterator it;
vector<int>str[500010];
int n, used[500010], ans;
bool cmp(int a, int b)
{
    return a > b;
}
int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j * j <= i; j++)
        {
            if(i % j == 0)
               {
                    str[i].push_back(j);
                    if(j * j != i)
                        str[i].push_back(i / j);
               }
        }
        sort(str[i].begin(), str[i].end(), cmp);
    }
    ans = 0;
    for(int a = 1; a < n; a++)
    {
        for(int x = 1; x * a < n; x++)
            {
             int yb = n - a * x;
             for(it = str[yb].begin(); it != str[yb].end(); it++)
                {
                    if((*it) <= a)
                        break;
                    if(used[*it] != a)
                    {
                        ans++;
                        used[*it] = a;
                    }
                }
            }
    }
    printf("%d\n", ans);
    return 0;
}