题目链接:https://www.lydsy.com/JudgeOnline/problem.php?id=2038
取袜子总的方案数为(r−l+1)∗(r−l)/2,取到两只同颜色的袜子的方案数sum[i]∗(sum[i]−1)/2的累加,sum[i]为每种颜色袜子的数量。
套下莫队的模板就好。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
ll n, m, block, te;
const int maxn = 5e5+5;
int a[maxn], num[maxn];
int ans[maxn][2];
struct node{
ll l, r, id;
}q[maxn];
ll gcd(ll a, ll b)
{
while(b > 0)
{
ll tmp = b;
b = a%b;
a = tmp;
}
return a;
}
ll C(ll x)
{
return x*(x-1)/2;
}
bool cmp(const node &s, const node &s1)
{
if(s.l/block == s1.l/block) return s.r < s1.r;
return s.l < s1.l;
}
void add(int x)
{
te -= C(num[a[x]]);
num[a[x]]++;
te += C(num[a[x]]);
}
void del(int x)
{
te -= C(num[a[x]]);
num[a[x]]--;
te += C(num[a[x]]);
}
int main()
{
ios::sync_with_stdio(false);
while(cin >> n >> m)
{
memset(num, 0, sizeof num);
block = ceil(sqrt(n));
for(int i = 1; i <= n; i++)
cin >> a[i];
for(int i = 1; i <= m; i++)
{
cin >> q[i].l >> q[i].r;
q[i].id = i;
}
sort(q+1, q+1+m, cmp);
int l = 1, r = 0;
te = 0;
for(int i = 1; i <= m; i++)
{
while(l < q[i].l) del(l++);
while(l > q[i].l) add(--l);
while(r < q[i].r) add(++r);
while(r > q[i].r) del(r--);
ll t = C(q[i].r-q[i].l+1);
ll g = gcd(t, te);
ans[q[i].id][0] = te/g;
ans[q[i].id][1] = t/g;
}
for(int i = 1; i <= m; i++)
printf("%d/%d\n", ans[i][0], ans[i][1]);
}
return 0;
}