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HDOJ 5147 Sequence II 树状数组
树状数组:
维护每一个数前面比它小的数的个数,和这个数后面比他大的数的个数
再枚举每个位置组合一下
Sequence II
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 121 Accepted Submission(s): 58
Problem Description
Long long ago, there is a sequence A with length n. All numbers in this sequence is no smaller than 1 and no bigger than n, and all numbers are different in this sequence.
Please calculate how many quad (a,b,c,d) satisfy:
1.1≤a<b<c<d≤n
2.Aa<Ab
3.Ac<Ad
Please calculate how many quad (a,b,c,d) satisfy:
1.
2.
3.
Input
The first line contains a single integer T, indicating the number of test cases.
Each test case begins with a line contains an integer n.
The next line follows n integersA1,A2,…,An .
[Technical Specification]
1 <= T <= 100
1 <= n <= 50000
1 <=Ai <= n
Each test case begins with a line contains an integer n.
The next line follows n integers
[Technical Specification]
1 <= T <= 100
1 <= n <= 50000
1 <=
Output
For each case output one line contains a integer,the number of quad.
Sample Input
1 5 1 3 2 4 5
Sample Output
4
Source
BestCoder Round #23
/* *********************************************** Author :CKboss Created Time :2014年12月20日 星期六 21时38分00秒 File Name :HDOJ5147.cpp ************************************************ */ #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <string> #include <cmath> #include <cstdlib> #include <vector> #include <queue> #include <set> #include <map> using namespace std; typedef long long int LL; const int maxn=55000; int a[maxn]; int n; LL sum1[maxn],sum2[maxn]; int t1[maxn],t2[maxn]; int lowbit(int x) { return x&(-x); } /// 1 找比当前数小的 2 找比当前数大的 void init() { memset(sum1,0,sizeof(sum1)); memset(sum2,0,sizeof(sum2)); memset(t1,0,sizeof(t1)); memset(t2,0,sizeof(t2)); } void add(int kind,int p) { if(kind==1) for(int i=p;i<maxn;i+=lowbit(i)) t1[i]+=1; else if(kind==2) for(int i=p;i;i-=lowbit(i)) t2[i]+=1; } int sum(int kind,int p) { int ret=0; if(kind==1) for(int i=p;i;i-=lowbit(i)) ret+=t1[i]; else if(kind==2) for(int i=p;i<maxn;i+=lowbit(i)) ret+=t2[i]; return ret; } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int T_T; scanf("%d",&T_T); while(T_T--) { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",a+i); init(); /// from left to right for(int i=1;i<=n;i++) { int ss=sum(1,a[i]); sum1[i]=ss; add(1,a[i]); } /// from right to left for(int i=n;i>=1;i--) { int ss=sum(2,a[i]); sum2[i]=sum2[i+1]+ss; add(2,a[i]); } LL ans=0; for(int i=2;i<=n-1;i++) { /// X...i i+1...X ans+=sum1[i]*sum2[i+1]; } cout<<ans<<endl; } return 0; }
HDOJ 5147 Sequence II 树状数组
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