N <tex2html_verbatim_mark>(3N
20000) <tex2html_verbatim_mark>ping pong players
live along a west-east street(consider the street as a line segment). Each
player has a unique skill rank. To improve their skill rank, they often compete
with each other. If two players want to compete, they must choose a referee
among other ping pong players and hold the game in the referee‘s house. For some
reason, the contestants can‘t choose a referee whose skill rank is higher or
lower than both of theirs. The contestants have to walk to the referee‘s house,
and because they are lazy, they want to make their total walking distance no
more than the distance between their houses. Of course all players live in
different houses and the position of their houses are all different. If the
referee or any of the two contestants is different, we call two games different.
Now is the problem: how many different games can be held in this ping pong
street?
Input
The first line of the input contains an integer T <tex2html_verbatim_mark>(1T
20) <tex2html_verbatim_mark>, indicating the number
of test cases, followed by T <tex2html_verbatim_mark>lines each of
which describes a test case.
Every test case consists of N +
1 <tex2html_verbatim_mark>integers. The first integer
is N <tex2html_verbatim_mark>, the number
of players. Then N <tex2html_verbatim_mark>distinct
integers a1, a2...aN <tex2html_verbatim_mark>follow,
indicating the skill rank of each player, in the order of west to east
( 1ai
100000 <tex2html_verbatim_mark>, i =
1...N <tex2html_verbatim_mark>).
Output
For each test case, output a single line contains an integer, the total number of different games.
Sample Input
1 3 1 2 3
Sample Output
1
树状数组问题
考虑第i个人当裁判时，假设a1到a(i-1)中有Ci个人比ai小，那么就有(i-1)-Ci个比ai大；同理，假设a(i+1)到an中有Di个比ai小，那么就有(n-i)-Di个比ai大。这样就可以算出，i
当裁判时有Ci*(n-i-Di)+(i-Ci-1)*Di种比赛。
这样就把问题转化为了求数组Ci和Di的问题：
C[i]可以这样求：
从左到右扫描所有的ai，令x[j]表示目前为止已经考虑的所有ai中，是否存在一个ai=j（x[j]=0表示不存在，x[j]=1表示存在），则Ci就是前缀和x[1]+x[2]+…+x[ai-1]。初始所有x[i]=0，在计算Ci时，需要先令x[ai]=1，即执行数状数组的加操作，然后求前缀和。这样即可求出数组C[i]。
D[i]的求法与C[i]同理，写代码的时候只要让for偱环倒过来进行即可
注意最终求和时要用long long保存，用int会溢出（此处WA两次…）
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 5 using namespace std; 6 7 int n; 8 int a[20050],temp[100050],c[20050],d[20050]; 9 10 int lowbit(int x) 11 { 12 return x&(-x); 13 } 14 15 int sum(int x) 16 { 17 int ret=0; 18 19 while(x>0) 20 { 21 ret+=temp[x]; 22 x-=lowbit(x); 23 } 24 25 return ret; 26 } 27 28 void add(int x,int t) 29 { 30 while(x<=100000) 31 { 32 temp[x]+=t; 33 x+=lowbit(x); 34 } 35 } 36 37 int main() 38 { 39 int kase; 40 41 scanf("%d",&kase); 42 43 while(kase--) 44 { 45 memset(c,0,sizeof(c)); 46 memset(d,0,sizeof(d)); 47 48 scanf("%d",&n); 49 50 for(int i=1;i<=n;i++) 51 scanf("%d",&a[i]); 52 53 memset(temp,0,sizeof(temp)); 54 for(int i=1;i<=n;i++) 55 { 56 add(a[i],1); 57 c[i]=sum(a[i]-1); 58 } 59 60 memset(temp,0,sizeof(temp)); 61 for(int i=n;i>=1;i--) 62 { 63 add(a[i],1); 64 d[i]=sum(a[i]-1); 65 } 66 67 long long ans=0; 68 for(int i=2;i<n;i++) 69 ans+=c[i]*(n-i-d[i])+d[i]*(i-1-c[i]); 70 71 printf("%lld\n",ans); 72 } 73 74 return 0; 75 }