题干：
Avin is studying how to synthesize data. Given an integer n, he constructs an interval using the following method: he first generates a integer r between 1 and n (both inclusive) uniform-randomly, and then generates another integer l between 1 and r (both inclusive) uniform-randomly. The interval [l, r] is then constructed. Avin has constructed two intervals using the method above. He asks you what the probability that two intervals intersect is. You should print p* q(−1)q(−1)(MOD 1, 000, 000, 007), while pqpq denoting the probability.
Input
Just one line contains the number n (1 ≤ n ≤ 1, 000, 000).
Output
Print the answer.
Sample Input
1 2
Sample Output
1 750000006
题目大意：
给定长度为n的区间，定义产生一个区间[l,r]的方式：第一次在[1,n]内等概率选一个点当做r，然后在[1,r]内等概率选一个点当l。
如此产生两个区间，问这两个区间相交（应该是指的有交点）的概率是多大？
解题报告：
对右端点分成三种情况，因为是古典改性直接用乘法原理计算对应事件发生的概率；累加即可。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
const ll mod = 1e9 + 7;
int n;
ll qpow(ll a,ll b){ll res=1;while(b) {if(b&1) res=res*a%mod;a=a*a%mod;b>>=1;}return res;
}
int main()
{while(~scanf("%d",&n)){printf("%lld\n",(n+1)*qpow(2*n,mod-2)%mod);}return 0;
}
)
