题目描述

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样例

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include[HTML_REMOVED]

include[HTML_REMOVED]

using namespace std;
typedef long long LL;
const int N = 1e5 + 10;
int a[N], b[N], c[N];
int main()
{
int n;
scanf(“%d”, &n);
for (int i = 0; i < n; i) scanf(“%d”, &a[i]);
for (int i = 0; i < n; i) scanf(“%d”, &b[i]);
for (int i = 0; i < n; i) scanf(“%d”, &c[i]);
sort(a, a + n); //二分需要满足单调性
sort(b, b + n);
sort(c, c + n);
LL res = 0; //答案可能会很大,会爆int
for (int i = 0; i < n; i)
{
int l = 0, r = n - 1; //二分查找a数组中最后一个小于b[i]的数的下标
while (l < r)
{
int mid = (l + r + 1) / 2;
if (a[mid] < b[i]) l = mid;
else r = mid - 1;
}
if (a[l] >= b[i]) //如果未找到小于b[i]的数,将x标记为-1,后续计算时 x+1==0
{
l = -1;
}
int x = l;
l = 0, r = n - 1;
while (l < r)
{
int mid = (l + r) / 2;
if (c[mid] > b[i]) r = mid;
else l = mid + 1;
}
if (c[l] <= b[i]) //如果未找到大于b[i]的数,将y标记为n,后续计算时 n-y==0;
{
r = n;
}
int y = r;
res += (LL)(x + 1)*(n - y);
}
printf(“%lld\n”, res);
return 0;
}

算法1

(暴力枚举) $O(n^2)$

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时间复杂度

参考文献

C++ 代码

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算法2

(暴力枚举) $O(n^2)$

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时间复杂度

参考文献

C++ 代码

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