UESTC Training for Data Structures——A

 

Problem  A

Problem Description

  The Romans have attacked again,This time they are much more than the Persians but Shapue is ready to defeat them.He says:"A lion is never afraid of a hundred sheep".
  Nevertheless Shapur has to find weaknesses in the Roman army to defeat them,So he gives the army a weakness number.In Shapur's opinion the weakness of an army is equal to the number of triplets i,j,k such that i<j<k and ai>aj>ak,where ax is the power of man standing at position x.The Roman army has one special trail -- powers of all the people in it are distinct.
  Help Shapur find out how weak the Romans are.

Input

The first line is an integer T(T<=10)--The number of test cases.
For each of the test case,the first line contains a single number n(3<=n<=10^6)--the number of men in Roman army.The next line contains N different possitive integers ai(1<=ai<=10^6)--power of men in the Romen army.

Output

For each of the T cases,output a single number,the weakness of the Roman army.

Sample Input

333 2 132 3 1410 8 3 1

Sample Output

104

/*将原来的数据按照从小到大一次hash成1--n，然后每次计算[1,i]中比a[i]小的元素的个数x每次ans增加的值就是 (i-1-x)*(a[i]-1-x)，最后输出ans的值即可*/#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<ctime>#define N 1000006using namespace std;struct data{    int num,id;} b[N];int a[N];int sum[N];void qsort(int st,int en)  //对输入的数据进行排序{    srand((unsigned)time(NULL));    //产生一个st到en的随机数 ll    int ll=rand()%(en-st)+st;    int i=st,j=en;    b[0]=b[ll]; b[ll]=b[st]; b[st]=b[0];    while(i<j)    {        while(i<j && b[0].num<=b[j].num) j--;        if(i<j) { b[i]=b[j]; i++; }        while(i<j && b[0].num>=b[i].num) i++;        if(i<j) { b[j]=b[i]; j--; }    }    b[i]=b[0];    if(st<i-1) qsort(st,i-1);    if(i+1<en) qsort(i+1,en);}__int64 get_sum(int *a,int pos)  //求和{    __int64 temp=0;    while(pos>0)    {        temp+=a[pos];        pos-=pos&(-pos);    }    return temp;}void update(int *a,int pos,int k)  //更新{    while(pos<N)    {        a[pos]+=k;        pos+=pos&(-pos);    }}int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        memset(sum,0,sizeof(sum));        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            b[i].num=a[i];            b[i].id=i;        }        qsort(1,n);        for(int i=1;i<=n;i++) a[b[i].id]=i; //将原来的数据按照从小到大hash成1--n        __int64 ans=0;        for(int i=1;i<=n;i++)        {            __int64 x=get_sum(sum,a[i]);  //找到[1,i]中，比a[i]小的元素的个数            ans+=(i-1-x)*(a[i]-1-x);            update(sum,a[i],1);  //将a[i]的元素的个数增加1个        }        printf("%I64d\n",ans);    }    return 0;}