Problem:
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 A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i.

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 \leq N \leq 10^5

1 \leq A_i \leq 10^9(1\leq i\leq N)

1 \leq B_i \leq 10^9(1\leq i\leq N)

1 \leq C_i \leq 10^9(1\leq i\leq N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N
B_1 ... B_N
C_1 ... C_N

Output

Print the number of different altars that Ringo can build.

Sample Input 1

2
1 5
2 4
3 6

Sample Output 1

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

Sample Input 2

3
1 1 1
2 2 2
3 3 3

Sample Output 2

27

Sample Input 3

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

Sample Output 3

87

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

//AtCoder Beginner Contest 077 C - Snuke Festival
//計算時間短縮その４
//下記修正済		
//miman = n - result + 1; //計算ミス
//miman = n - result + 1;

//#define DEBUG1
//#define DEBUG2
//#define DEBUG3

#include <stdio.h>
#include <stdlib.h>

int P[3][100000];

//変数
int miman = -1;
int koeru = -1;

int cmp(const void *p, const void *q) {
	return *(int*)p - *(int*)q;
}

int main(void) {
	int n = 0;
	int i, j;

	//入力パラメータ


	//入力作業
	scanf("%d", &n);
	for (i = 0; i < 3; i++) {
		for (j = 0; j < n; j++) {
			scanf("%d", &P[i][j]);
		}
	}
#ifdef DEBUG
	for (i = 0; i < 3; i++) {
		for (j = 0; j < n; j++) {
			printf("P[%d][%d] = %d\n", i, j, P[i][j]);
		}
	}
#endif
	//ソート作業
	//for (i = 0; i < 3; i++) {
	//	qsort(P[i], n, sizeof(int), cmp);//[1]のソートは不要なので削除
	//}
		qsort(P[0], n, sizeof(int), cmp);
		qsort(P[2], n, sizeof(int), cmp);
	
#ifdef DEBUG2
	for (i = 0; i < 3; i++) {
		for (j = 0; j < n; j++) {
			printf("P[%d][%d] = %d\n", i, j, P[i][j]);
		}
	}
#endif

#ifdef DEBUG2
	//テストパターン
	//tansaku(P[0], n, P[1][0]);
	//tansaku(P[2], n, P[1][0]);
	//tansaku(P[0], n, P[2][5]);
	//tansaku(P[1], n, P[0][0]);
#endif

#ifdef DEBUG2
	//printf("mid_r = %d\n", mid_r);
	//printf("mid_l = %d\n", mid_l);
	//getch();
#endif

	int r, t;
	int total = 0;

	for (j = 0; j < n; j++) {
		miman = 0; //変数リセット
		koeru = 0;//変数リセット
		int tmp = -1;
		tansaku(P[0], n, P[1][j]);
		tmp = miman;
		tansaku(P[2], n, P[1][j]);
		tmp = tmp * koeru;
		total = total + tmp;
#ifdef DEBUG3
		printf("total = %d\n", total);
#endif
	}
	printf("%d\n", total);
	//getch();
#ifdef DEBUG3
	getch();
#endif
	return 0;
}


//入力パラメータ
//Vac>0 一致無し 端では無い
//int a[6] = { 1,2,4,6,8,9 };//ソート済の配列　要素数は仮
//int n = 6;//要素数
//int x = 7;//探したい値

		  //Vac>0 一致無し　＋オーバー
		  //int a[6] = { 1,2,4,6,8,9 };//ソート済の配列　要素数は仮
		  //int n = 6;//要素数
		  //int x = 10;//探したい値

		  //Vac>0　一致
		  //int a[6] = { 1,2,3,4,9,10 };//ソート済の配列　要素数は仮
		  //int n = 6;//要素数
		  //int x = 9;//探したい値

		  //Vac>0 一致無し　-オーバーはあり得ないので検討不要

		  //Vac<0 一致無し
		  //int a[6] = { 1,3,4,5,10,11 };//ソート済の配列　要素数は仮
		  //int n = 6;//要素数
		  //int x = 2;//探したい値

		  //Vac<0 一致無し
		  //int a[6] = { 2,3,4,5,10,11 };//ソート済の配列　要素数は仮
		  //int n = 6;//要素数
		  //int x = 1;//探したい値
int tansaku(int a[], int n, int x) {

	int left = 0;
	int right = n - 1;
	int result = -1;
	int mid = -1;
	int _L = -1;
	int	_R = -1;
	//int miman = -1;
	//int koeru = -1;

	//正常処理
#ifdef DEBUG2
	printf("x = %d\n", x);
#endif
	int Vec = 0;//ベクトル指標

	while (left <= right) {
		mid = (left + right) / 2;

		if (a[mid] == x) {
#ifdef DEBUG2
			printf("a[%d]=%d == %d\n", mid, a[mid], x);
#endif
			result = mid;
			Vec = 0;
			left = right;//処理終了フラグ
		}
		if (a[mid] > x) {
#ifdef DEBUG2
			printf("a[%d]=%d > %d\n", mid, a[mid], x);
#endif
			right = mid - 1;
			Vec = -1;
		}
		if (a[mid] < x) {
#ifdef DEBUG2
			printf("a[%d]=%d < %d\n", mid, a[mid], x);
#endif
			left = mid + 1;
			Vec = 1;
		}
	}
	//一致の場合
	if (result != -1) {
#ifdef DEBUG2
		printf("Vec = %d\n", Vec);
		printf("result = %d\n", result);
		printf("P[%d] = x \n", result, x);
#endif
	}

	//一致では無い場合
	if (result == -1) {
		//場所の調査
		if (Vec > 0) {
#ifdef DEBUG2
			printf("Vec = %d\n", Vec);
#endif
			_L = mid;
#ifdef DEBUG2
			printf("P[%d] < %d \n", _L, x);
#endif
		}
		if (Vec < 0) {
#ifdef DEBUG2
			printf("Vec = %d\n", Vec);
#endif
			_R = mid;
#ifdef DEBUG2
			printf("%d < P[%d]\n", x, _R);
#endif
		}
	}

	//未満と超えるの要素数の算出
	if (result != -1) {
		//miman = n - result + 1; //計算ミス
		miman = n - result + 1;
		koeru = n - miman - 1;
	}
	if (_L != -1 && _R == -1) {
		miman = _L + 1;
		koeru = n - miman;
	}
	if (_L == -1 && _R != -1) {
		miman = _R + 1 - 1;
		koeru = n - miman;
	}
#ifdef DEBUG2
	printf("未満要素数 = %d\n", miman);
	printf("超える要素数 = %d\n", koeru);
	getch();
#endif
	return 0;
}