Codeforces 261 d

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Solution report: a sequence of A1, A2, a3 ......... an, f (I, j, x) AK is equal to the number of X (I <= k <= J), so that I <J, determine how many pairs of I and j make f (1, I, AI)> F (J, N, AJ ).

Scan this sequence from left to right. num1 [I] is equal to the number of [I] equals to, similarly, you can obtain num2 [I] by scanning from right to left, maintain the number of num2 [I] scanned from right to left in a tree array, and add all the values to the tree array first, then, scan num1 [I] from the left to the right to determine the number of num1 [I] In num2 each time, at the same time, after judgment, we need to subtract 1 from the number of num2 [I], because the position of num1 [I] has exceeded the position of num2, therefore, num2 [I] cannot be used to calculate the number in the future. It should be deleted. The purpose of using a tree array is to quickly determine how many num2 [I] are smaller than num1 [I], so as to achieve).

 1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<map> 5 #include<algorithm> 6 using namespace std; 7 #define LL long long 8 #define maxn 1000005 9 LL n,tot;10 map<LL,LL> mp1,mp2;11 LL a[maxn],num1[maxn],num2[maxn],num3[maxn];12 LL tree[maxn];13 LL find(LL d,int l,int r)14 {15     while(l < r)16     {17         int mid = (l + r) / 2;18         if(d <= num1[mid]) r = mid;19         else l = mid + 1;20     }21     if(num1[l] != d) return l - 1;22     else return l;23 }24 void insert(int l,int d)25 {26     for(int i = l;i <= n;i += (-i & i))27     tree[i] += d;28 }29 LL sum(int l)30 {31     LL tot = 0;32     for(int i = l;i > 0;i -= (-i & i))33     tot += tree[i];34     return tot;35 }36 37 int main()38 {39     40     while(scanf("%lld",&n)!=EOF)41     {42         for(int i = 1;i <= n;++i)43         scanf("%lld",&a[i]);44         memset(tree,0,sizeof(tree));45         memset(num1,0,sizeof(num1));46         memset(num2,0,sizeof(num2));47         mp1.clear();48         mp2.clear();49         for(int i  = 1;i <= n;++i)50         {51             if(mp1.insert(pair<LL,LL> (a[i],1)).second == 1)52             num1[i] = 1;53             else54             {55                 mp1[a[i]] = mp1[a[i]] + 1;56                 num1[i] = mp1[a[i]];57             }58         }59         for(int i = n;i >= 1;--i)60         {61             if(mp2.insert(pair<LL,LL> (a[i],1)).second == 1)62             num2[i] = 1;63             else64             {65                 mp2[a[i]] = mp2[a[i]] + 1;66                 num2[i] = mp2[a[i]];67             }68         }69         for(int i = 1;i <= n;++i)70         insert(num2[i],1);71         tot = 0;72         for(int i = 1;i <= n;++i)73         {74             insert(num2[i],-1);75             tot += sum(num1[i]-1);76         }77         printf("%lld\n",tot);78     }79     return 0;80 }

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