bzoj2038

题解：

题目要求输出分数

那么分母就是n*n

分子就是总数=相同颜色^2

然后用莫队来做

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=50005;
int n,m,Be[N],ans,sum[N],col[N];
struct Mo
{
    int l,r,id;
    ll A,B;
}q[N];
ll S(ll x)
{
    return x*x;
}
ll gcd(ll x,ll y)
{
    if (!y)return x;
    return gcd(y,x%y);
}
int cmp(Mo a,Mo b)
{
    if (Be[a.l]==Be[b.l])return a.r<b.r;
    return a.l<b.l;
}
int cmp1(Mo a,Mo b)
{
    return a.id<b.id;
}
void revise(int x,int add)
{
    ans-=S(sum[col[x]]);
    sum[col[x]]+=add;
    ans+=S(sum[col[x]]);
}
int main()
{
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++)
     {
         scanf("%d",&col[i]);
         Be[i]=i/(int)sqrt(n)+1;
     }
    for (int i=1;i<=m;i++)
     {
         scanf("%d%d",&q[i].l,&q[i].r);
         q[i].id=i;
     } 
    sort(q+1,q+m+1,cmp);
    int l=1,r=0;
    for (int i=1;i<=m;i++)
     {
         while (l<q[i].l)revise(l,-1),l++;
         while (l>q[i].l)revise(l-1,1),l--;
         while (r<q[i].r)revise(r+1,1),r++;
         while (r>q[i].r)revise(r,-1),r--;
         if (q[i].l==q[i].r)
          {
              q[i].A=0;
              q[i].B=1;
              continue;
          }
         q[i].A=ans-(q[i].r-q[i].l+1);
        q[i].B=1LL*(q[i].r-q[i].l+1)*(q[i].r-q[i].l);
        ll gc=gcd(q[i].A,q[i].B);
        q[i].A/=gc;
        q[i].B/=gc;
     } 
    sort(q+1,q+m+1,cmp1);
    for (int i=1;i<=m;i++)printf("%lld/%lld\n",q[i].A,q[i].B); 
}