hdu5147 (sequence 2) 树状数组

 1 #include <iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #include<vector>
 6 #include<map>
 7 using namespace std;
 8 #define INF 0x7fffffff
 9 #define MOD 1000000007
10 
11 
12 #define N 100010
13 #define MAX 1000100
14 
15 int C[MAX], S[MAX], b[N],d[N];
16 long long f[N];
17 long long total[N], ans;
18 int num[N], T, s, t, i, j;
19 
20 int Lowbit(int x){
21     return x&(x^(x-1));
22 }
23 
24 void add(int pos,int num,int *P) {
25     while(pos <= MAX) {
26         P[pos] += num;
27         pos += Lowbit(pos);
28     }
29 }
30 
31 int Sum(int end,int *P) {
32     int cnt = 0;
33     while(end > 0) {
34         cnt += P[end];
35         end -= Lowbit(end);
36     }
37     return cnt;
38 }
39 
40 void init(){
41     total[0] = 0;
42     for(i = 1; i < N; ++i){
43         total[i] = total[i-1] + i;
44     }
45 }
46 
47 int main() {
48     //init();
49     int t;
50     cin>>t;
51 
52     while(t--) {
53 
54         scanf("%d",&T);
55 
56         memset(C,0,sizeof(C));
57         memset(S,0,sizeof(S));
58         memset(b,0,sizeof b);
59         memset( d,0,sizeof d);
60         memset(f,0,sizeof f);
61         memset(num,0,sizeof num);
62         //memset(num,0,sizeof(num));
63         //memset(b,0,sizeof(b));
64         //ans = 0;
65         for(j = 0; j < T; j ++) {//因为第一个数前面比它小的数没有，所以j要从0开始
66             scanf("%d",&num[j]);
67             add(num[j]+1,1,C);
68             b[j] = Sum(num[j], C);//Sum(num[j],C)求的就是小于s的个数，j - Sum(num[j],C)就是前j个数中大于num[j]的个数
69             //printf("%d ",b[j]);
70         }
71         //printf("\n");
72         //ans = 0;
73         for(j = T-1; j > -1; --j){//反过来求第j个数右边中大于它的数的个数。
74             add(num[j]+1 ,1, S);
75             d[j] =T-1-j-Sum(num[j] ,S);//Sum(num[j],S)求的就是小于num[j]的个数
76             //b[j] -= Sum(num[j]+1,S) - Sum(num[j],S)-1;
77             //printf("%d ",b[j]);
78             //ans += total[b[j]];
79         }
80         f[T-1]=0;
81        for(int j=T-2;j>=0;j--)
82         f[j]=d[j+1]+f[j+1];
83        long long int res=0;
84        for(int j=0;j<T;j++)
85         res+=b[j]*f[j];
86        cout<<res<<endl;
87 
88     }
89     return 0;
90 }
91 
92 代码君