这一题，设每个区间[l,r]各个颜色的袜子数分别为a,b,c,d,…a,b,c,d,…，每个区间[l,r]的答案就是(C2a+C2b+C2c+C2d+⋯)/C2r−l+1(Ca2+Cb2+Cc2+Cd2+⋯)/Cr−l+12，展开化简得：
(a2+b2+c2+d2+⋯−a−b−c−d−⋯)/((r−l+1)∗(r−l+1−1))(a2+b2+c2+d2+⋯−a−b−c−d−⋯)/((r−l+1)∗(r−l+1−1))
(a2+b2+c2+d2+⋯−(r−l+1))/((r−l+1)∗(r−l))(a2+b2+c2+d2+⋯−(r−l+1))/((r−l+1)∗(r−l))
其中(a2+b2+c2+d2+⋯)(a2+b2+c2+d2+⋯)便可作为莫队算法处理的区间答案，开个数组记录abcd...的个数可以在O(1)转移到相邻区间。
另外特判区间l=r的情况。。
/// .-~~~~~~~~~-._ _.-~~~~~~~~~-.
/// __.' ~. .~ `.__
/// .'// \./ \\`.
/// .'// | \\`.
/// .'// .-~"""""""~~~~-._ | _,-~~~~"""""""~-. \\`.
/// .'//.-" `-. | .-' "-.\\`.
/// .'//______.============-.. \ | / ..-============.______\\`.
/// .'______________________________\|/______________________________`.
//#pragma GCC optimize("Ofast")
//#pragma comment(linker, "/STACK:102400000,102400000")
//#pragma GCC target(sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx)
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;
#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e6+5;
const int maxx=1e6+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}
inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}
void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;
int n,m;
int belong[maxn],a[maxn],block;
LL Ans=0;
LL fz[maxn],fm[maxn],ans[maxn];
struct node {
int l,r,id;
}Q[maxn];
bool cmp(node a,node b) {
if(belong[a.l]==belong[b.l]) {
return a.r<b.r;
}
return belong[a.l]<belong[b.l];
}
void build(){
block=sqrt(n);
FOR(1,n,i)
belong[i]=(i-1)/block+1;
}
void add(int x) {
Ans+=ans[a[x]];
ans[a[x]]++;
}
void del(int x) {
ans[a[x]]--;
Ans-=ans[a[x]];
}
void solve() {
s_2(n,m);
FOR(1,n,i) s_1(a[i]);
FOR(1,m,i) {
s_2(Q[i].l,Q[i].r);
Q[i].id=i;
}
sort(Q+1,Q+m+1,cmp);
build();
int L=1,R=0;
FOR(1,m,i) {
W(L<Q[i].l) {
del(L);
L++;
}
W(L>Q[i].l) {
L--;
add(L);
}
W(R<Q[i].r) {
R++;
add(R);
}
W(R>Q[i].r) {
del(R);
R--;
}
LL k=((1LL)*(Q[i].r-Q[i].l+1))*(Q[i].r-Q[i].l)/2;
if(Ans!=0) {
LL g=__gcd(Ans,k);
fz[Q[i].id]=Ans/g;
fm[Q[i].id]=k/g;
}
else {
fz[Q[i].id]=Ans;
fm[Q[i].id]=1;
}
}
FOR(1,m,i)
printf("%lld/%lld\n",fz[i],fm[i]);
}
int main() {
//freopen( "1.in" , "r" , stdin );
//freopen( "1.out" , "w" , stdout );
int t=1;
//init();
//s_1(t);
for(int cas=1;cas<=t;cas++) {
//printf("Case #%d: ",cas);
solve();
}
}