[Bzoj 1303] [CQOI2009] Median figure "0.0"

Title Link: BZOJ-1303

Problem analysis

First, locate position Pos for B, and then assign the value of less than B in the sequence to-1, and the value greater than B is assigned to 1.

Expand from B to the left, keep counting Sum[i, b-1], and then add one to Cnt[sum[i, b-1], so that the Sum of each left is the number of values.

then expand from B to the right, counting Sum[b + 1, i], then Ans + = cnt[sum[b + 1, I]].

Code

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath > #include <algorithm> using namespace std; const int MAXN = 100000 + 5; int n, b, num, Pos, Sum;int A[MAXN], CNTL[MAXN * 2]; typedef long Long LL; LL Ans; int main () {     scanf ("%d%d", &n, &b);    for (int i = 1; I <= n; ++i) {        scanf ("%d", &num);        if (num = = b) {            a[i] = 0;            Pos = i;            Continue;        }        if (num < b) A[i] =-1;        else a[i] = 1;    }    Sum = 0;    for (int i = Pos; I >= 1; i.) {        Sum + = A[i];        ++cntl[sum + n];    }    Sum = 0; Ans = 0LL;    for (int i = Pos; I <= n; ++i) {        Sum + = A[i];        Ans + = (LL) cntl[-sum + n];    }    printf ("%lld\n", Ans);    return 0;}

　　

[Bzoj 1303] [CQOI2009] Median figure "0.0"