# 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__()```
```

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## Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

## Direct Connections

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.) do

## Subsequence Weighting

A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence. You are given a sequence A in which every element is a pair of integers i.e A = [(a1, w1), (a2, w2),..., (aN, wN)]. For a subseqence B = [(b1, v1), (b2, v2), ...., (bM, vM)] of the given sequence : We call it increasing if for every i (1 <= i < M ) , bi < bi+1. Weight(B) =