题目要求的是平均值不小于K的,那么可以将问题变成,对所有的都减去K,然后求“权值和大于等于0”的子串的个数有多少个?
于是,我们可以求,以每个点作为子串结尾的点时候的可能的子串的数量,这里就可以用前缀和来维护了,然后加上前缀和小于等于当前前缀和的点的个数,就是答案了。
用一个主席树维护一下即可。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e5 + 7;
const ll _UP = 2e9 + 1;
int N, K, a[maxN];
int tot = 0, root = 0, tree[maxN * 40], lc[maxN * 40], rc[maxN * 40];
void Insert(int &rt, ll l, ll r, ll qx)
{
if(!rt) rt = ++tot;
tree[rt]++;
if(l == r) return;
ll mid = HalF;
if(qx <= mid) Insert(lc[rt], l, mid, qx);
else Insert(rc[rt], mid + 1, r, qx);
}
int query(int rt, ll l, ll r, ll ql, ll qr)
{
if(!rt) return 0;
if(ql <= l && qr >= r) return tree[rt];
ll mid = HalF;
if(qr <= mid) return query(lc[rt], l, mid, ql, qr);
else if(ql > mid) return query(rc[rt], mid + 1, r, ql, qr);
else return query(lc[rt], l, mid, ql, qr) + query(rc[rt], mid + 1, r, ql, qr);
}
signed main()
{
scanf("%d%d", &N, &K);
for(int i=1; i<=N; i++) { scanf("%d", &a[i]); a[i] -= K; }
ll ans = 0;
Insert(root, -_UP, _UP, 0);
for(int i=1, sum = 0; i<=N; i++)
{
sum += a[i];
ans += query(root, -_UP, _UP, -_UP, sum);
Insert(root, -_UP, _UP, sum);
}
printf("%lld\n", ans);
return 0;
}