题目传送门。
（多倍经验：P1471，P10511，P5142）
首先做查询。平均数好做，考虑方差怎么搞。大力推柿子：
\[\begin{align*}
s^2&=\frac{\sum\limits_{i=1}^n(x_i-\overline{x})^2}{n}\\
&=\frac{\sum\limits_{i=1}^n(x_i^2-2x_i+\overline{x}^2)}{n}\\
&=\frac{(n\overline{x}^2)+(\sum\limits_{i=1}^nx_i^2)-2\overline{x}\cdot(\sum\limits_{i=1}^nx_i)}{n}\\
&=\frac{\overline{x}(n\overline{x}-2\cdot\sum\limits_{i=1}^nx_i)+(\sum\limits_{i=1}^nx_i^2)}{n}\\
&=\frac{(\sum\limits_{i=1}^nx_i^2)-\overline{x}\cdot(\sum\limits_{i=1}^nx_i)}{n}\\
&=\frac{(\sum\limits_{i=1}^nx_i^2)-\frac{(\sum\limits_{i=1}^nx_i)^2}{n}}{n}\\
&=\frac{n\cdot(\sum\limits_{i=1}^nx_i^2)-(\sum\limits_{i=1}^nx_i)^2}{n}
\end{align*}
\]
这样就容易维护了。接下来考虑区间加时怎么更新整块的平方和，依旧推柿子：
\[\begin{align*}
\sum\limits_{i=1}^n(x_i+v)^2&=\sum\limits_{i=1}^n(x_i^2-2x_iv+v^2)\\
&=(\sum\limits_{i=1}^nx_i^2)+2v\cdot(\sum\limits_{i=1}^nx_i)+nv^2
\end{align*}
\]
然后分个块就可以做了。代码：
#include<iostream>
#include<cmath>
#define ll long long
using namespace std;const int N=1e5+5;
const int M=320;int n,m;ll a[N];namespace OIfast{#define gc getchar()inline unsigned read(){unsigned n=0;char c=gc;while(!isdigit(c))c=gc;while(isdigit(c))n=(n<<3)+(n<<1)+(c^48),c=gc;return n;}}using namespace OIfast;namespace Math{inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}inline ll pf(ll x){return x*x;}}using namespace Math;namespace FK{int k,tot;short loc[N];int L[M],R[M];ll s[M],ss[M],la[M];inline void pushdown(int i){for(int j=L[i];j<=R[i];++j)a[j]+=la[i];return la[i]=0,void();}inline void pushup(int i){s[i]=ss[i]=0;for(int j=L[i];j<=R[i];++j)s[i]+=a[j],ss[i]+=pf(a[j]);return ;}inline void init(){k=sqrt(n),tot=ceil((1.0*n)/(1.0*k));for(int i=1;i<=tot;++i){L[i]=(i-1)*k+1,R[i]=min(n,L[i]+k-1);for(int j=L[i];j<=R[i];++j)loc[j]=i;pushup(i);}return ;}inline void upd(int l,int r,ll v){if(loc[l]==loc[r]){pushdown(loc[l]);for(int i=l;i<=r;++i)a[i]+=v;pushup(loc[l]);}else{pushdown(loc[l]);for(int i=l;i<=R[loc[l]];++i)a[i]+=v;pushup(loc[l]);pushdown(loc[r]);for(int i=L[loc[r]];i<=r;++i)a[i]+=v;pushup(loc[r]);for(int i=loc[l]+1;i<=loc[r]-1;++i)la[i]+=v,ss[i]+=2*v*s[i]+(R[i]-L[i]+1)*pf(v),s[i]+=(R[i]-L[i]+1)*v;}return ;}inline ll qry_s(int l,int r){ll res=0;if(loc[l]==loc[r]){pushdown(loc[l]);for(int i=l;i<=r;++i)res+=a[i];pushup(loc[l]);}else{pushdown(loc[l]);for(int i=l;i<=R[loc[l]];++i)res+=a[i];pushup(loc[l]);pushdown(loc[r]);for(int i=L[loc[r]];i<=r;++i)res+=a[i];pushup(loc[r]);for(int i=loc[l]+1;i<=loc[r]-1;++i)res+=s[i];}return res;}inline ll qry_ss(int l,int r){ll res=0;if(loc[l]==loc[r]){pushdown(loc[l]);for(int i=l;i<=r;++i)res+=pf(a[i]);pushup(loc[l]);}else{pushdown(loc[l]);for(int i=l;i<=R[loc[l]];++i)res+=pf(a[i]);pushup(loc[l]);pushdown(loc[r]);for(int i=L[loc[r]];i<=r;++i)res+=pf(a[i]);pushup(loc[r]);for(int i=loc[l]+1;i<=loc[r]-1;++i)res+=ss[i];}return res;}#define pii pair<ll,ll>#define mp make_pair#define fir first#define sec secondinline pii dx(int l,int r){ll a=qry_s(l,r),b=r-l+1;if(!a)return mp(0,1);ll g=gcd(a,b);return mp(a/g,b/g);}inline pii fc(int l,int r){ll a=(r-l+1)*qry_ss(l,r)-pf(qry_s(l,r)),b=pf(r-l+1);if(!a)return mp(0,1);ll g=gcd(a,b);return mp(a/g,b/g);}}using namespace FK;inline void work(){int op=read(),l=read(),r=read();if(1==2)puts("wow");else if(op==1){ll d=read();upd(l,r,d);}else if(op==2){pii tmp=dx(l,r);printf("%lld/%lld\n",tmp.fir,tmp.sec);}else if(op==3){pii tmp=fc(l,r);printf("%lld/%lld\n",tmp.fir,tmp.sec);}return ;
}signed main(){n=read(),m=read();for(int i=1;i<=n;++i)a[i]=read();init();while(m--)work();return 0;
}
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