Codeforces #261 D

Codeforces #261 D

D. Pashmak and Parmida's problem

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Parmida is a clever girl and she wants to participate in Olympiads this year. Of course she wants her partner to be clever too (although he's not)! Parmida has prepared the following test problem for Pashmak.

There is a sequence a that consists of n integers a1, a2, ..., an. Let's denote f(l, r, x) the number of indices k such that: l ≤ k ≤ r andak = x. His task is to calculate the number of pairs of indicies i, j (1 ≤ i < j ≤ n) such that f(1, i, ai) > f(j, n, aj).

Help Pashmak with the test.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 106). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print a single integer — the answer to the problem.

 

简单的树状数组求逆序数。。。开始怎么也没想到。。。

答案没用long long wa一次

  1 #include <cstring>
  2 
  3 #include <iostream>
  4 
  5 #include <algorithm>
  6 
  7 #include <cstdio>
  8 
  9 #include <cmath>
 10 
 11 #include <map>
 12 
 13 #include <cstdlib>
 14 
 15 #define M(a,b) memset(a,b,sizeof(a))
 16 
 17 using namespace std;
 18 
 19 
 20 
 21 int n;
 22 
 23 int num[1000006];
 24 
 25 int savefro[1000006],saveb[1000006];
 26 
 27 int ans[1000006];
 28 
 29 int flag[1000006];
 30 
 31 map<int,int> front,back;
 32 
 33 
 34 
 35 int bit[2000006],cnn;
 36 
 37 
 38 
 39 int sum(int i)
 40 
 41 {
 42 
 43     int res = 0;
 44 
 45     while(i>0)
 46 
 47     {
 48 
 49         res+=bit[i];
 50 
 51         i-=i&-i;
 52 
 53     }
 54 
 55     return res;
 56 
 57 }
 58 
 59 
 60 
 61 void add(int i,int x)
 62 
 63 {
 64 
 65     while(i<=cnn)
 66 
 67     {
 68 
 69         bit[i]+=x;
 70 
 71         i+=i&-i;
 72 
 73     }
 74 
 75 }
 76 
 77 
 78 
 79 int main()
 80 
 81 {
 82 
 83     while(scanf("%d",&n)==1)
 84 
 85     {
 86 
 87         front.clear();
 88 
 89         back.clear();
 90 
 91         M(saveb,0);
 92 
 93         M(savefro,0);
 94 
 95         M(ans,0);
 96 
 97         M(flag,0);
 98 
 99         M(bit,0);
100 
101         long long res = 0;
102 
103         for(int i = 0;i<n;i++)
104 
105         {
106 
107             scanf("%d",&num[i]);
108 
109         }
110 
111         for(int i = 0;i<n;i++)
112 
113         {
114 
115             savefro[i] = ++front[num[i]];
116 
117         }
118 
119         cnn = n;
120 
121         for(int i = n-1;i>=0;i--)
122 
123         {
124 
125             saveb[i] = ++back[num[i]];
126 
127             add(saveb[i],1);
128 
129             res+=sum(savefro[i-1]-1);
130 
131            // cout<<' '<<sum(savefro[i-1]-1)<<' '<<savefro[i-1]<<endl;
132 
133         }
134 
135 
136 
137         /*cout<<"savefro"<<endl;
138 
139         for(int i = 0;i<n;i++)
140 
141         {
142 
143             cout<<savefro[i]<<' ';
144 
145         }
146 
147         cout<<endl;
148 
149 
150 
151         cout<<"saveb"<<endl;
152 
153         for(int i = 0;i<n;i++)
154 
155         {
156 
157             cout<<saveb[i]<<' ';
158 
159         }
160 
161         cout<<endl;
162 
163 
164 
165         cout<<"flag"<<endl;
166 
167         for(int i = 0;i<n;i++)
168 
169         {
170 
171             cout<<flag[i]<<' ';
172 
173         }
174 
175         cout<<endl;*/
176 
177         printf("%I64d\n",res);
178 
179     }
180 
181     return 0;
182 
183 }

 

posted @ 2014-10-19 19:46  haohaooo  阅读(208)  评论(0)  编辑  收藏  举报