[Tree array] HDU 4000

1 ~ N sequence, and how many groups of x <z <Y

Method: Tree array, first obtain the number of xyz + xzy, very simple, calculate the number larger than AI, the number smaller than Ai, this is achieved using a tree array, at the beginning, I thought it was done backwards, I .e., subscript n ~ 1. If the sequence is not necessary, you can find a number P smaller than that of AI, and calculate it on the draft paper, calculate the number larger than AI (n-I-a [I] + p) {calculate this subprocess: The number smaller than AI is P, it is reasonable to say that there is a small number of ai-1 than Ai, then there must be a ai-1-p smaller than AI in the right side of the AI, occupying a ai-1-p position, then the right side of AI is left N-(I + 1) space, so the right side is greater than the number of ai n-(I + 1)-(ai-1-p) = (n-I-a [I] + p). Note that my subscript starts from 0}. After finding The subscripts that are larger than AI and smaller than Ai, you can calculate them.

#include <map>#include <set>#include <list>#include <queue>#include <deque>#include <stack>#include <string>#include <time.h>#include <cstdio>#include <math.h>#include <iomanip>#include <cstdlib>#include <limits.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;#define LL long long#define PI acos(-1.0)#define FRE freopen("a.txt","r",stdin)#define MAX INT_MAX#define MIN INT_MIN#define eps 1e-10#define MOD 100000007#define N  100005LL c[N];int a[N];int lowbit(int t){    return t&(-t);}LL sum(int x){    LL tot=0;    while(x){        tot+=c[x];        x-=lowbit(x);    }    return tot;}void add(int x){    while(x<N){        c[x]+=1;        x+=lowbit(x);    }}int main(){FRE;    int t;    scanf("%d",&t);    int ca;    for(ca=1;ca<=t;ca++){        //memset(c,0,sizeof(c));        int n;        scanf("%d",&n);        int i,j;        for(i=0;i<=n;i++)c[i]=0;        for(i=0;i<n;i++)scanf("%d",&a[i]);        LL s1=0,s2=0;        for(i=0;i<n;i++){            LL p=sum(a[i]);//the number less than a[i]            s2+=(n-i-a[i]+p)*p;//x<y<z            s1+=(n-i-a[i]+p)*(n-i-a[i]+p-1)/2;            add(a[i]);        }        printf("Case #%d: %I64d\n",ca,(s1-s2)%MOD);    }    return 0;}