题目描述
题目分析
60分解法:
直接暴力循环每一个数进行比较
cpp
#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
typedef long long ll;
ll n, a[N], b[N], c[N], ans;
int main()
{
cin >> n;
for(int i = 1; i <= n; i ++)cin >> a[i];
for(int i = 1; i <= n; i ++)cin >> b[i];
for(int i = 1; i <= n; i ++)cin >> c[i];
for(int i = 1; i <= n; i ++)
{
for(int j = 1; j <= n; j ++)
{
for(int k = 1; k <= n; k ++)
{
if(a[i] < b[j] && b[j] < c[k])ans ++;
}
}
}
cout << ans;
return 0;
}
满分解法:
由于ABC的值是完全独立的所以可以使用乘法原理
由B作为一个判断点,看有多少个A符合要求,再看有多少个C符合要求,最后的答案则为两部分相乘的结果
cpp
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
ll n, a[N], b[N], c[N], s[N], cnt[N];
ll sa[N];//sa[i]表示在a[]中有多少个数小于b[i]
ll sc[N];//sc[i]表示在c[]中有多少个数大于b[i]
int main()
{
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> n;
for(int i = 1; i <= n; i ++)cin >> a[i], a[i] ++;
for(int i = 1; i <= n; i ++)cin >> b[i], b[i] ++;
for(int i = 1; i <= n; i ++)cin >> c[i], c[i] ++;
//求sa[]
for(int i = 1; i <= n; i ++)cnt[a[i]] ++;
for(int i = 1; i <= N; i ++)s[i] = s[i - 1] + cnt[i];
for(int i = 1; i <= n; i ++)sa[i] = s[b[i] - 1];
//求sc[]
memset(cnt, 0, sizeof cnt);
memset(s, 0, sizeof s);
for(int i = 1; i <= n; i ++)cnt[c[i]] ++;
for(int i = 1; i <= N; i ++)s[i] = s[i - 1] + cnt[i];
for(int i = 1; i <= n; i ++)sc[i] = s[N] - s[b[i]];
//枚举每个b[i]
ll ans = 0;
for(int i = 1; i <= n; i ++)ans += sa[i] * sc[i];
cout << ans;
return 0;
}