P2717 寒假作业
平均值不小于 $k$ ，将所有的数都减去 $k$ ，等价于平均值不小于 $0$ ，即和不小于 $0$ 。
考虑 cdq ，对于 mid 左右两侧，分别求后缀和、前缀和。排序后做双指针，可以做到 $O(n \log ^ 2 n)$ 。
另有加强版 P7868 。
查看代码
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| #include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write(Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar(x % 10 + '0');
}
const int N = 1e5 + 10;
int n, m, w[N], tmp[N];
LL cdq(int l, int r)
{
if (l == r)
return w[l] >= 0;
int mid = l + r >> 1;
LL res = cdq(l, mid) + cdq(mid + 1, r);
tmp[mid] = w[mid], tmp[mid + 1] = w[mid + 1];
for (int i = mid - 1; i >= l; i--)
tmp[i] = tmp[i + 1] + w[i];
for (int i = mid + 2; i <= r; i++)
tmp[i] = tmp[i - 1] + w[i];
sort(tmp + l, tmp + mid + 1);
sort(tmp + mid + 1, tmp + r + 1);
for (int i = l, j = r; i <= mid; i++)
{
while (j > mid && tmp[i] + tmp[j] >= 0)
j--;
res += r - j;
}
return res;
}
int main()
{
read(n), read(m);
for (int i = 1; i <= n; i++)
read(w[i]), w[i] -= m;
write(cdq(1, n));
return 0;
}
|