codevs1516 平均分数--逆序对

设sum[i]为i的前缀和 
 
显然对于左边的我们可以通过求逆序对，解决了 
这只是大于等于左边的，那么右边呢 
看这里 
 
嗯，那么就是关于l我们要 求一个非严格逆序对（记作a）对于r求一个严格逆序对(记作b) 
那么，由数轴可知，ans = a - b; 
嗯 
然后就可以无脑求了 
不过要记得sum[0]，否则就会错过只有一个人的情况 
嗯，然后总方案数就是高斯巧解…… 
然后约分，gcd也好

#include <iostream>

#include <cstdio>

#include <algorithm>

#include <cstring>

#define KILL puts("haha");

usingnamespacestd;

constlonglongMAXN = 1000000+ 5;

longlongnum[MAXN],zh[MAXN],ans;

bool flag;

void merge(longlongl,longlongr)

{

   longlongmid = (l + r) >> 1;

   longlongs = l,t = mid + 1;

   longlongtot = l;

   while(s<= mid && t <= r)

   {

        if(num[s]< num[t] || ((num[s] == num[t]) * flag))

            zh[tot++] = num[s++];

        else

            zh[tot++] = num[t++],ans += mid - s+ 1;

   }

   while(s<= mid)

        zh[tot++] = num[s++];

   while(t<= r)

        zh[tot++] = num[t++];

   for(longlongi = l;i <= r;i ++)

        num[i] = zh[i];

   return;

}

void merge_sort(longlongl,longlongr)

{

   if(l>= r)

        return;

   longlongmid = (l + r) >> 1;

   merge_sort(l,mid);

   merge_sort(mid+1,r);

   merge(l,r);

   return;

}

longlong gcd(longlongx,longlongy)

{

   returny?gcd(y,x % y):x;

}

longlong gs(longlongx)

{

   return(x * (x + 1))>> 1;

}

longlong n,l,r;

longlong a[MAXN];

longlong sum[MAXN];

longlong ansa,ansb;

int main()

{

   scanf("%lld%lld %lld",&n,&l,&r);

   for(longlongi = 1;i<= n;i ++)

   {

        scanf("%lld",&a[i]);

        sum[i] = sum[i - 1] + a[i];

   }

   for(longlongi = 0;i<= n;i ++)

        num[i] = l * i - sum[i];

   ans = 0;

   merge_sort(0,n);

   ansa = ans;

   flag = true;

   ans = 0;

   for(longlongi = 0;i<= n;i ++)

        num[i] = r * i - sum[i];

   merge_sort(0,n);

   ansb = ans;

   ans = ansa - ansb;

   longlongmo = gs(n);

   longlongk = gcd(ans,mo);

   ans /= k;

   mo /= k;

   if(ans== 0)

       puts("0");

   elseif(ans- mo == 0)

        puts("1");

   else

        printf("%lld/%lld",ans,mo);

   return0;

}