HDU 7100 Cut The Wire
Problem Description
Each streetlight x adheres to specific connection rules:
Kris stands midway between street lights numbered n and n+1, capable of severing every wire along his path.
Specifically, he will cut all wires connecting any two lights a and b where a \leq n and b > n.
Curious about the count, he seeks your assistance in figuring it out.
Each test case begins with an integer T (1 \leq T \leq 105), denoting the number of test cases to follow.
Following this are T lines, each containing an integer n within the range [1, 109].
For each test case, output a single integer representing the solution.
2 12 60
Sample Output
10 50
题意
给定点x，则若其编号是偶数，则连接至x/2；同样地，在这种情况下也连接至3*x + 1。
设给定一个节点编号为n，则克里斯位于节点n与节点n+1之间，并切断所有连接到节点n或节点n+1的边。
问题在于询问会被切断的所有边的数量是多少。
不管n是取到偶数值还是奇数值，在这个区间内的所有偶数值点之间的连线都将全部被切除；当且仅当n取到的是一个奇数值时，则切除这样的连线数量将等于\frac{n - \frac{n - 1}{3}}{2}条；而如果此时n是一个偶数值，则同样数量即\frac{n - \frac{n - 1}{3}}{2}条。
AC代码：
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int n,m;
int main()
{
int t;
cin>>t;
while(t--)
{
cin>>n;
int res=n-n/2;
if(n%2)
res+=(n-(n-1)/3+1)/2;
else
res+=(n-(n-1)/3)/2;
cout<<res<<endl;
}
}
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