# Direct Connections

### Problem Statement :

```Enter-View ( EV )  is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points.

Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) doesn't know anything about the modern telecom technologies, except for peer-to-peer connections. Even worse, his thoughts on peer-to-peer connections are extremely faulty: he believes that, if Pi  people are living in city i , there must be at least  cables from city  to every other city of EV - this way he can guarantee no congestion will ever occur!

Mr. Treat hires you to find out how much cable they need to implement this telecommunication system, given the coordination of the cities and their respective population.

Note that The connections between the cities can be shared. Look at the example for the detailed explanation.

Input Format

A number T is given in the first line and then comes T blocks, each representing a scenario.

Each scenario consists of three lines. The first line indicates the number of cities (N). The second line indicates the coordinates of the N cities. The third line contains the population of each of the cities. The cities needn't be in increasing order in the input.

Output Format

For each scenario of the input, write the length of cable needed in a single line modulo 1, 000, 000, 007.

Constraints

1  <=  T <=  20
1  <=  N  <= 200,000
Pi  <= 10,000

Border to border length of the country  <=  1,  000, 000, 000```

### Solution :

```                            ```Solution in C :

In C ++ :

#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <string>
#include <string.h>
#define pb push_back
#define mp make_pair
#define SS(a,b) scanf("%d%d",&a,&b);
#define S(a) scanf("%d",&a);
#define SSL(a,b) scanf("%lld%lld",&a,&b);
#define SL(a) scanf("%lld",&a);
#define SSS(a,b,c) scanf("%d %d %d",&a,&b,&c);
#define GI ({int t;scanf("%d",&t);t;})
#define GL ({ll t;scanf("%lld",&t);t;})
#define MAXN 500000
#define FOR(i,a,n) for(int i=a;i<n;i++)
#define REP(i,n) FOR(i,0,n)
#define INPUT freopen("input.txt","r",stdin);
#define OUTPUT freopen("output1.txt","w",stdout);
#define disvec(v) { for(int vec_index=0;vec_index<v.size();vec_index++) cout<<v[vec_index]<<" "; cout<<endl;}
using namespace std;
typedef  long long LL;
typedef  long long ll;
pair<LL,LL>input[200100];
map<LL,int>m;
LL DistBIT[200100];
LL SumBIT[200100];
LL MOD=1000000007LL;
int maxval=200010;
void init(){
REP(i,200100)DistBIT[i]=SumBIT[i]=0;
}
void update_Dist(int idx,LL val){
while(idx<=maxval){
DistBIT[idx]=(DistBIT[idx]+val);
idx+=(idx & -idx);
}
}
void update_Sum(int idx,LL val){
while(idx<=maxval){
SumBIT[idx]=(SumBIT[idx]+val)%MOD;
idx+=(idx & -idx);
}
}
LL query_cnt(int idx){
LL res=0;
while(idx>0){
res=(res+SumBIT[idx]);
idx-=(idx & -idx);
}
return res;
}
LL QUERY_cnt(int start,int end){
if(start>end)return 0;
return (query_cnt(end)-query_cnt(start-1));
}
LL query_Dist(int idx){
LL res=0;
while(idx>0){
res=(res+DistBIT[idx]);
idx-=(idx & -idx);
}
return res;
}
LL QUERY_Dist(int start, int end){
if(start>end)return 0;
return (query_Dist(end)-query_Dist(start-1));
}
pair<LL,LL>Mapp[200100];
int main(){
int t=GI;
while(t--){
init();
int n=GI;
m.clear();
REP(i,n){scanf("%lld",&input[i].second);}
REP(i,n){scanf("%lld",&input[i].first);}
REP(i,n){
Mapp[i].first=input[i].second;
Mapp[i].second=input[i].first;
}
sort(Mapp,Mapp+n);
int total=2;
REP(i,n)m[Mapp[i].first]=total,total++;

sort(input,input+n);

LL ans=0;
update_Dist(m[input[0].second],input[0].second);
update_Sum(m[input[0].second],1);
for(int i=1;i<n;i++){
LL xi=input[i].second;
LL curmapped=m[input[i].second];

LL Leftcnt=QUERY_cnt(1,curmapped-1);
LL Rightcnt=QUERY_cnt(curmapped+1,total+2);

LL LeftDist=QUERY_Dist(1,curmapped-1);
LL RightDist=QUERY_Dist(curmapped+1,total+2);

LL Tempans=xi*(Leftcnt-Rightcnt);

Tempans-=LeftDist;
Tempans+=RightDist;
Tempans=(Tempans*input[i].first)%MOD;

ans=(ans+Tempans)%MOD;
update_Dist(curmapped,xi);
update_Sum(curmapped,1);
}
cout<<(ans+MOD)%MOD<<endl;
}
GI;
return 0;
}

In Java :

import java.io.IOException;
import java.util.Arrays;
import java.util.Comparator;
import java.util.StringTokenizer;

public class DirectConnection {
static final class IO {
//Standard IO
static StringTokenizer tokenizer = null;

static int ni() {
return Integer.parseInt(ns());
}

static long nl() {
return Long.parseLong(ns());
}

static double nd() {
return Double.parseDouble(ns());
}

static String ns() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
} catch (IOException e) {
}
}
}

static String nline() {
tokenizer = null;
String ret = null;
try {
} finally {
return ret;
}
}
}

static final class City {
int i;
int cordinate;
int cable;
}

static class FenwickTree {
static long MOD = 1000000000 + 7;

public static void update(long[] arr, int pos, int val)
{
int len = arr.length;
for (; pos < len; pos |= (pos + 1))
arr[pos] = (arr[pos] + val + MOD) % MOD;
}

/** Function to query **/
public static long query(long[] arr, int pos)
{
long sum = 0;
for (; pos >= 0; pos = (pos & (pos + 1)) - 1)
sum = (arr[pos] + sum + MOD) % MOD;

return sum;
}
}

static long f(City[] cityCor) {
Arrays.sort(cityCor, new Comparator<City>() {
@Override
public int compare(City o1, City o2) {
return Integer.compare(o1.cordinate, o2.cordinate);
}
});
long[] iList = new long[cityCor.length];
long[] cord = new long[cityCor.length];
long[] cable = new long[cityCor.length];
for (int i = 0; i < cityCor.length; i++) {
cityCor[i].i = i;
FenwickTree.update(cord, i, cityCor[i].cordinate);
FenwickTree.update(cable, i, cityCor[i].cable);
FenwickTree.update(iList, i, 1);
}
Arrays.sort(cityCor, new Comparator<City>() {
@Override
public int compare(City o1, City o2) {
return Integer.compare(o2.cable, o1.cable);
}
});
long sum = 0;
for (City c : cityCor) {
sum = (sum + g(c, iList, cord)) % FenwickTree.MOD;
//System.out.println(c.cable + "--" + sum);
FenwickTree.update(iList, c.i, -1);
FenwickTree.update(cord, c.i, -1 * c.cordinate);

}
return sum;
}

private static long g(City c, long[] iList, long[] cord) {
long MOD = FenwickTree.MOD;
int i = c.i;
long l = FenwickTree.query(cord, i);
long n = FenwickTree.query(cord, cord.length - 1);
n = (n - l + FenwickTree.MOD) % FenwickTree.MOD;
//System.out.println(i + "&& " + l + "." + n);
long li = FenwickTree.query(iList, i);
long ni = FenwickTree.query(iList, iList.length - 1);
ni = (ni - li + FenwickTree.MOD) % FenwickTree.MOD;
//System.out.println(i + "|| " + li + "." + ni);

long cc = (((c.cordinate * li - l + MOD) % MOD) * c.cable) % MOD;
long nn = (((n - c.cordinate * ni + MOD) % MOD) * c.cable) % MOD;
return (cc + nn) % MOD;
}
public static void main(String[] argv) {
int T = IO.ni();
for (int t = 0; t < T; t++) {
int N = IO.ni();
City[] citArr = new City[N];

for (int i = 0; i < N; i++) {
citArr[i] = new City();
citArr[i].cordinate = IO.ni();
}
for (int i = 0; i < N; i++) {
citArr[i].cable = IO.ni();

}
System.out.println(f(citArr));
}
}
}

In C :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
struct node {
long int c, p;
};
struct node a[200000];
#define MOD(x) ((x) < 0 ? -(x) : (x))
#define MODULO(x) ((x) > 1000000007 ? (x) % 1000000007 : (x))
int cmp(void *a, void *b)
{
return ((struct node *)b)->p - ((struct node *)a)->p;
}
int main() {
long int n, t, i, j, tot, ans;
scanf("%ld", &t);
while (t--) {
scanf("%ld", &n);
tot = 0;
for(i = 0; i < n; i++) {
scanf("%ld", &a[i].c);
//tot += c;
}
for(i = 0; i < n; i++) {
scanf("%ld", &a[i].p);
}
qsort(a, n, sizeof(a[0]), (__compar_fn_t)cmp);
ans = 0;
for(i = 0; i < n; i++) {
tot = 0;
for(j = i+1; j < n; j++)
tot += MOD(a[j].c - a[i].c);
ans += MODULO(tot) * MODULO(a[i].p);
ans = MODULO(ans);
}
printf("%ld\n", ans);

}
return 0;
}

In Python3 :

class fenpiece:
__slots__ = ['x','p','px','c']
def __init__(self,x=0,p=0,px=0,c=0):
self.x = x
self.p = p
self.px = px
self.c = c
self.x += other.x
self.p += other.p
self.px += other.px
self.c += other.c
return self
return fenpiece(self.x,self.p,self.px,self.c)
def __sub__(self,other):
return fenpiece(self.x-other.x,self.p-other.p,self.px-other.px,self.c-other.c)

def fensum(seq,i):
sum = 0
while i:
sum += seq[i-1]
i -= i&-i
return sum

def fensumrange(seq,i,j):
return fensum(seq,j) - fensum(seq,i)

i += 1
bound = len(seq) + 1
while i < bound:
seq[i-1] += v
i += i&-i

pBound = 10001
magicmod = 1000000007
fenlist = [fenpiece() for i in range(pBound)]
T = int(input())
for t in range(T):
total = 0
N = int(input())
X = [int(s) for s in input().split()]
P = [int(s) for s in input().split()]
cities = sorted(zip(X,P))
cable = 0
for x,p in cities:
underP = fensum(fenlist,p)
overP = fensumrange(fenlist,p,pBound)
cable =  (cable + p*(underP.c*x - underP.x) + overP.p*x - overP.px)%magicmod
print(cable)
for f in fenlist:f.__init__()```
```

## Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

## Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

## Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink