Fun with Integers -codeforces

阿新 • • 發佈：2018-11-16

讀懂題意就好，有個最直接的想法，列舉\(a,b\)，複雜度\(O(n^2)\)。然而換成列舉\(a,x\),複雜度\(O(nlg(n))\)。\(<a,b>\)和\(<b,a>\)的貢獻只能算一次。

程式碼

#include <bits/stdc++.h>
using namespace std;

int n;
long long ans;

int main()
{
    cin >> n;
    for (int a = 2; a <= n / 2; ++a) for (int x = 2; x * a <= n; ++x) ans += x * 4;
    cout << ans << endl;

    return 0;
}

Fun with Integers -codeforces

讀懂題意就好，有個最直接的想法，列舉\(a,b\)，複雜度\(O(n^2)\)。然而換成列舉\(a,x\),複雜度\(O(nlg(n))\)。\(<a,b>\)和\(<b,a>\)的貢獻只能算一次。程式碼 #include <bits/stdc++.h> usin

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Fun with Integers -codeforces

程式碼

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