大家好，又见面了，我是全栈君，今天给大家准备了Idea注册码。
Number sequence
Given a number sequence which has N element(s), please calculate the number of different collocation for three number Ai, Aj, Ak, which satisfy that Ai < Aj > Ak and i < j < k.
Input
The first line is an integer N (N <= 50000). The second line contains N integer(s): A1, A2, …, An(0 <= Ai <= 32768).
Output
There is only one number, which is the the number of different collocation.
Sample Input
5 1 2 3 4 1
Sample Output
6
这题能够用树状数组做，开两个一维的树状数组分别记录当前点前面的比这点小的个数和后面比这点大的个数。
#include<iostream>#include<stdio.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<queue>#include<stack>#include<string>#include<algorithm>using namespace std;#define maxn 50005#define ll long longint b1[maxn],b2[maxn],a[maxn];int lowbit(int x){ return x&(-x);}void update1(int pos,int num){ while(pos<=maxn){ b1[pos]+=num;pos+=lowbit(pos); }}int getsum1(int pos){ int num=0; while(pos>0){ num+=b1[pos];pos-=lowbit(pos); } return num;}void update2(int pos,int num){ while(pos<=maxn){ b2[pos]+=num;pos+=lowbit(pos); }}int getsum2(int pos){ int num=0; while(pos>0){ num+=b2[pos];pos-=lowbit(pos); } return num;}int main(){ int n,m,i,j; ll num=0; while(scanf("%d",&n)!=EOF) { memset(b1,0,sizeof(b1)); memset(b2,0,sizeof(b2)); for(i=1;i<=n;i++){ scanf("%d",&a[i]); a[i]++; if(i==1){ update1(a[i],1);continue; } update2(a[i],1); } num=0; for(i=2;i<=n-1;i++){ num+=getsum1(a[i]-1)*getsum2(a[i]-1); update1(a[i],1); update2(a[i],-1); } printf("%lld\n",num); } return 0;}版权声明：本文内容由互联网用户自发贡献，该文观点仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 举报，一经查实，本站将立刻删除。
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