D
问题描述
一个无向图包含 2020 条边,如果图中没有自环和重边,请问最少包含多少个结点?
答案提交
这是一道结果填空的题,你只需要算出结果后提交即可。本题的结果为一个整数,在提交答案时只填写这个整数,填写多余的内容将无法得分。
#include <bits/stdc++.h>
using namespace std;
int main()
{
for(int i = 1;i<=100;i++)//代数进去算,大概n个节点的话
{ //最多可以有(n*(n-1))/2条边
if((i*(i-1))/2>=2020)
{
cout<<i<<endl;
return 0;
}
}
return 0;
}
答案是:65

题A: https://blog.youkuaiyun.com/FG_future/article/details/114456485
题B:https://blog.youkuaiyun.com/FG_future/article/details/114456603
题C:https://blog.youkuaiyun.com/FG_future/article/details/114456686
题D:https://blog.youkuaiyun.com/FG_future/article/details/114735667
题E:https://blog.youkuaiyun.com/FG_future/article/details/114735794
题F:https://blog.youkuaiyun.com/FG_future/article/details/114735934
题G:https://blog.youkuaiyun.com/FG_future/article/details/114736107
题H:https://blog.youkuaiyun.com/FG_future/article/details/114736189
题J:https://blog.youkuaiyun.com/FG_future/article/details/114736383
本文探讨了一个无向图在不含自环和重边的情况下,若边的数量为2020时,至少需要多少个结点。通过简单的数学计算得出结论,并提供了C++代码实现。
2504

被折叠的 条评论
为什么被折叠?



