#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define L(a) a<<1
#define R(a) a<<1|1
#define P pair<int,int>
#define pii pair<ll,ll>
#define lowbit(x) ((x)&-(x))
#define pi acos(-1.0)
#define gcd(x,y) __gcd(x,y)
#define unq(x) x.erase(unique(x.begin(), x.end()), x.end())
#define mem(x,k) memset(x,k,sizeof x)

const int dx[8]={0,0,1,-1,1,-1,1,-1};
const int dy[8]={1,-1,0,0,-1,1,1,-1};
const int inf = 0x3f3f3f3f;
const int mod =1e9+7;
// const ll inf=9e18;

const double eps = 1e-5;
const int maxn = 5e5+10;
const int N=1e8+10;
const double Pi=3.1415926535;


ll w[maxn];
int block;
ll sum[maxn];

struct node{
    int id=0;
    int l,r;
    bool operator<(const node &a)const{
        if(l/block==a.l/block){
            if((l/block)&1) return r<a.r;
            return r>a.r;
        }
        return l<a.l;
    }
}op[maxn];
ll ans[maxn];
ll b[maxn];
ll cnt;
void solve(ll &a,ll &b){
    if(a==0) return;
    int temp=gcd(a,b);
    a/=temp;
    b/=temp;
}

void add(int x){
    sum[x]++;
    cnt+=sum[x]-1;
}
void del(int x){
    cnt-=sum[x]-1;
    sum[x]--;
}
int main() {
    int n,m;
    cin>>n>>m;
    block=sqrt(n);
    for(int i=1;i<=n;i++) cin>>w[i];
    for(int i=1;i<=m;i++){
        int l,r;
        cin>>op[i].l>>op[i].r;
        op[i].id=i;
    }
    sort(op+1,op+1+m);
    int r=0,l=1;
    for(int i=1;i<=m;i++){
        int L=op[i].l,R=op[i].r;
        if(L==R){
            ans[op[i].id]=0;
            continue;
        }
        while(l<L) del(w[l++]);
        while(l>L) add(w[--l]);
        while(r<R) add(w[++r]); 
        while(r>R) del(w[r--]);
        ans[op[i].id]=cnt;
        b[op[i].id]=(op[i].r-op[i].l+1)*(op[i].r-op[i].l)/2;
    }
    for(int i=1;i<=m;i++){
        solve(ans[i],b[i]);
        if(ans[i]!=0) printf("%lld/%lld\n",ans[i],b[i]);
        else printf("0/1\n");
    }
}

我的意思是 如果如果当前答案为cnt 然后 a颜色 有k 个 如果再来一个a颜色 cnt+=k