題干：

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

題目大意：

給定長度為n的區間，定義產生一個區間[l,r]的方式：第一次在[1,n]內等概率選一個點當做r，然后在[1,r]內等概率選一個點當l。

如此產生兩個區間，問這兩個區間相交（應該是指的有交點）的概率是多大？

解題報告：

對右端點分成三種情況，因為是古典改性直接用乘法原理計算對應事件發生的概率；累加即可。

AC代碼：

#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; }

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