C - Snuke Festival

原题地址:http://abc077.contest.atcoder.jp/tasks/arc084_a;

Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower.

He has N parts for each of the three categories. The size of the i-th upper part is Ai, the size of the i-th middle part is Bi, and the size of the i-th lower part is Ci.

To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar.

How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.

Constraints

• 1≤N≤105
• 1≤Ai≤109(1≤i≤N)
• 1≤Bi≤109(1≤i≤N)
• 1≤Ci≤109(1≤i≤N)
• All input values are integers.

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Input

Input is given from Standard Input in the following format:

N
A1 … AN
B1 … BN
C1 … CN

Output

Print the number of different altars that Ringo can build.

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Sample Input 1

Copy

2
1 5
2 4
3 6

Sample Output 1

Copy

3

The following three altars can be built:

• Upper: 1-st part, Middle: 1-st part, Lower: 1-st part
• Upper: 1-st part, Middle: 1-st part, Lower: 2-nd part
• Upper: 1-st part, Middle: 2-nd part, Lower: 2-nd part

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Sample Input 2

Copy

3
1 1 1
2 2 2
3 3 3

Sample Output 2

Copy

27

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Sample Input 3

Copy

6
3 14 159 2 6 53
58 9 79 323 84 6
2643 383 2 79 50 288

Sample Output 3

Copy

87
题目意思:给三层数据,求从小到大的可能顺序的可能情况;
解题思路:没有太多难点,但是容易出错;

代码:

#include<iostream>
#include<string>
#include<algorithm>
#include <string.h>
#include <stdio.h>
#include <math.h>
#include <set>
#include <queue>
#include <stack>
#include <map>
 
using namespace std;
typedef long long LL;
const int INF = int(1e9);
const int MAX = 1000020;
LL N;
LL a[MAX],b[MAX],c[MAX];
 
void Cin(LL a[])
{
    for(int i = 1;i<=N;i++)
        scanf("%lld",&a[i]);
}
 
int main()
{
    cin>>N;
    Cin(a),Cin(b),Cin(c);
    sort(a+1,a+1+N),sort(b+1,b+1+N),sort(c+1,c+1+N);
    LL sum = 0;
    LL ai = 0,ci = 0;
    for(int i = 1;i<=N;i++){
        while(a[ai+1]<b[i]&&ai+1<=N)
            ai++;
        while(c[ci+1]<=b[i]&&ci+1<=N)
            ci++;
       // cout<<ai<<" "<<ci<<endl;
        sum += ai*(N-ci);  //因为相乘可能会溢出,所以要为LL;
    }
 
    cout<<sum<<endl;
 
    return 0;
}