bzoj 5055: 膜法师 树状数组+离散

先枚举每一个数，看它前面有几个比它小，算一下和为sum1，后面有几个比它大，算一下和为sum2，对答案的贡献为A[i]*sum1*sum2。

离散化后，树状数组就可以了。

就是倒着一边，顺着一边，统计就ok了，离散是为了大小，先后顺序（方便去重）

 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<map>
 #define mod 19260817
 #define N 300007
 #define ll long long
 using namespace std;

 int n,top;
 ll ans;
 ll a[N],tree[N],c[N];
 map<int,int>p;
 struct Node
 {
     int zhi,id;
 }b[N];

 bool cmp(Node x,Node y)
 {
     return x.zhi<y.zhi;
 }
 int lowbit(int x){return x&(-x);}
 ll query(int x)
 {
     ll res=;
     for (int i=x;i>=;i-=lowbit(i))
         res+=tree[i];
     res%=mod;
     return res;
 }
 void add(int x,ll z)
 {
     for (int i=x;i<=top;i+=lowbit(i))
         tree[i]=(tree[i]+z)%mod;
 }
 int main()
 {
     p.clear();
     scanf("%d",&n);
     for (int i=;i<=n;i++)
     {
         scanf("%d",&a[i]);
         a[i]%=mod;
         b[i].zhi=a[i];
         b[i].id=i;
     }
     sort(b+,b+n+,cmp);
     b[].zhi=-;
     for (int i=;i<=n;i++)
         if (b[i].zhi!=b[i-].zhi) p[b[i].zhi]=++top;
     for (int i=;i<=n;i++)
     {
         c[i]=1ll*query(p[a[i]]-)*a[i]%mod;
         add(p[a[i]],a[i]);
     }
     memset(tree,,sizeof(tree));
     for (int i=n;i>=;i--)
     {
         ans=(ans+1ll*c[i]*(query(top)-query(p[a[i]])+mod))%mod;
         add(p[a[i]],a[i]);
     }
     printf("%lld",ans);
 }