思路：树状数组。
考虑第i个人当裁判，那么只要计算出在他之前比他小的乘在他之后比他大的与在他之前比他大的乘在他之后比他小的，那么用两个树状数组维护一下就行了。复杂的（n*log(n)）
1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<string.h> 5 #include<queue> 6 #include<stack> 7 #include<math.h> 8 using namespace std; 9 typedef long long LL; 10 int bitl[200005]; 11 int bitr[200005]; 12 int lowbit(int x); 13 void addl(int x); 14 void addr(int x); 15 int askl(int x); 16 int askr(int x); 17 int ans[100005]; 18 int bns[100005]; 19 int aksl[200005]; 20 int aksr[200005]; 21 int main(void) 22 { 23 int T; 24 scanf("%d",&T); 25 while(T--) 26 { 27 int n; 28 scanf("%d",&n); 29 memset(bitl,0,sizeof(bitl)); 30 memset(bitr,0,sizeof(bitr)); 31 int i,j; 32 for(i = 1; i <= n; i++) 33 { 34 scanf("%d",&ans[i]); 35 } 36 int cn = 1; 37 for(i = n; i >= 1; i--) 38 { 39 bns[cn++] = ans[i]; 40 } 41 for(i = 1; i <= n; i++) 42 { 43 aksl[i] = askl(ans[i]-1); 44 addl(ans[i]); 45 aksr[i] = askr(bns[i]-1); 46 addr(bns[i]); 47 } 48 LL sum = 0; 49 for(i = 2;i <= n-1;i++) 50 { 51 sum = sum + (LL)aksl[i]*(LL)(n-i-aksr[n-i+1])+(LL)(i - 1 - aksl[i])*(LL)aksr[n-i+1]; 52 } 53 printf("%lld ",sum); 54 } 55 return 0; 56 } 57 int lowbit(int x) 58 { 59 return x&(-x); 60 } 61 void addl(int x) 62 { 63 while(x <= 100000) 64 { 65 bitl[x]++; 66 x += lowbit(x); 67 } 68 } 69 void addr(int x) 70 { 71 while(x <= 100000) 72 { 73 bitr[x]++; 74 x += lowbit(x); 75 } 76 } 77 int askl(int x) 78 { 79 int sum = 0; 80 while(x > 0) 81 { 82 sum += bitl[x]; 83 x -= lowbit(x); 84 } 85 return sum; 86 } 87 int askr(int x) 88 { 89 int sum = 0; 90 while(x > 0) 91 { 92 sum += bitr[x]; 93 x -= lowbit(x); 94 } 95 return sum; 96 }