Description：
有1到n这些数字各一个。你用这些数字进行若干轮游戏。 对于每一轮，如果剩下的数字个数超过1个，那么就等概率随机选择两个剩下的数字删去。如果这两个数字互质，得一分。 重复以上操作直到没数字可以删除为止。请问最后期望得多少分？
输入格式
一行一个整数n (1≤n≤5000)。
输出格式
输出一个最简约数
b
a
表示答案。
输入样例
2
4
输出样例
1/1
5/3
题意很简单，在
n
n
n 个数中选取
2
2
2 个一共有
n
∗
(
n
−
1
)
n*(n-1)
n∗(n−1) 种，我们直接暴力统计
n
n
n 内互质的一共有
c
n
t
cnt
cnt 对，那么概率就是
c
n
t
n
∗
(
n
−
1
)
\frac{cnt}{n*(n-1)}
n∗(n−1)cnt,再乘上总数量就是答案，但是要保证成对所以总数量为
(
n
/
2
)
∗
2
(n/2)*2
(n/2)∗2
AC代码：
#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", (n))
#define pdd(n, m) printf("%d %d\n", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, n, a) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
int ret = 0, sgn = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
sgn = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
{
ret = ret * 10 + ch - '0';
ch = getchar();
}
return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
if (a > 9)
Out(a / 10);
putchar(a % 10 + '0');
}
ll gcd(ll a, ll b)
{
return b == 0 ? a : gcd(b, a % b);
}
ll lcm(ll a, ll b)
{
return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
if (a >= mod)
a = a % mod + mod;
ll ans = 1;
while (b)
{
if (b & 1)
{
ans = ans * a;
if (ans >= mod)
ans = ans % mod + mod;
}
a *= a;
if (a >= mod)
a = a % mod + mod;
b >>= 1;
}
return ans;
}
// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
return qpow(a, p - 2, p);
}
///扩展欧几里得
ll exgcd(ll a, ll b, ll &x, ll &y)
{
if (b == 0)
{
x = 1, y = 0;
return a;
}
ll d = exgcd(b, a % b, x, y);
ll z = x;
x = y;
y = z - y * (a / b);
return d;
}
///中国剩余定理模板0
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
ll M = 1, y, x = 0;
for (int i = 0; i < n; ++i) //算出它们累乘的结果
M *= a[i];
for (int i = 0; i < n; ++i)
{
int w = M / a[i];
ll tx = 0;
ll t = exgcd(w, a[i], tx, y); //计算逆元
x = (x + w * (b[i] / t) * x) % M;
}
return (x + M) % M;
}
int main()
{
ll n, cnt = 0;
sld(n);
if (n == 1)
{
puts("0/1");
return 0;
}
for (ll i = 1; i <= n; i++)
{
for (ll j = i + 1; j <= n; j++)
{
if (gcd(i, j) == 1)
{
cnt++;
}
}
}
ll z = cnt * (n / 2) * 2;
ll m = n * (n - 1);
ll g = gcd(z, m);
z /= g;
m /= g;
printf("%lld/%lld", z, m);
return 0;
}
