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）
1 #include<cstdio> 2 #include<algorithm> 3 #include<cmath> 4 #include<iostream> 5 #include<cstring> 6 #include<map> 7 #define mod 19260817 8 #define N 300007 9 #define ll long long 10 using namespace std; 11 12 int n,top; 13 ll ans; 14 ll a[N],tree[N],c[N]; 15 map<int,int>p; 16 struct Node 17 { 18 int zhi,id; 19 }b[N]; 20 21 bool cmp(Node x,Node y) 22 { 23 return x.zhi<y.zhi; 24 } 25 int lowbit(int x){return x&(-x);} 26 ll query(int x) 27 { 28 ll res=0; 29 for (int i=x;i>=1;i-=lowbit(i)) 30 res+=tree[i]; 31 res%=mod; 32 return res; 33 } 34 void add(int x,ll z) 35 { 36 for (int i=x;i<=top;i+=lowbit(i)) 37 tree[i]=(tree[i]+z)%mod; 38 } 39 int main() 40 { 41 p.clear(); 42 scanf("%d",&n); 43 for (int i=1;i<=n;i++) 44 { 45 scanf("%d",&a[i]); 46 a[i]%=mod; 47 b[i].zhi=a[i]; 48 b[i].id=i; 49 } 50 sort(b+1,b+n+1,cmp); 51 b[0].zhi=-1000; 52 for (int i=1;i<=n;i++) 53 if (b[i].zhi!=b[i-1].zhi) p[b[i].zhi]=++top; 54 for (int i=1;i<=n;i++) 55 { 56 c[i]=1ll*query(p[a[i]]-1)*a[i]%mod; 57 add(p[a[i]],a[i]); 58 } 59 memset(tree,0,sizeof(tree)); 60 for (int i=n;i>=1;i--) 61 { 62 ans=(ans+1ll*c[i]*(query(top)-query(p[a[i]])+mod))%mod; 63 add(p[a[i]],a[i]); 64 } 65 printf("%lld",ans); 66 }