【樹狀數組】hdu 4000

給出一個1~N的序列，問有多少組x<z<y

方法：樹狀數組，先求出xyz+xzy的個數，很簡單，算出比ai大的個數，比ai小的個數，這個用樹狀數組實現，開始想的時候是倒過來建樹的，即下標n~1順序，發現沒必要，順序也可以，求出比ai小的個數p，然後草稿紙算算，算出比ai大的個數是(n-i-a[i]+p){求這個式子過程：比ai小的個數是p，按理來說有ai-1個比ai小的數，那麼肯定有ai-1-p個比ai小的在ai右面，佔據了ai-1-p個位置，然後ai右面剩下n-（i+1）個空位，所以右面比ai大的個數有n-(i+1)-(ai-1-p)=(n-i-a[i]+p)個，注意我的下標由0開始}，求出比ai大和比ai小後就可以接著算。

#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;}