题目链接 P2717 寒假作业

  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;
}