# Jaggu Playing with Balloons

### Problem Statement :

```Jaggu is a little kid and he likes playing with water balloons. He took 1 million ( 106 ) empty buckets and he filled the bucket with water balloons under the instruction of his sister Ishika.
His sister gives him two types of commands:

R pos1 pos2 which implies that jaggu needs to tell her what is the total number of water balloons in the bucket from pos1 to pos2 (both included).

U pos M plus which implies that he has to work like the function

Update(pos,M,plus)

void Update(int pos,int M,int plus)
{
int N=1000000;  //1 million
for (int i=1;i<=50;i++)
{
int back = pos
for(int j=1;j<=1000;j++)
{
add M water ballons at bucket pos
int s,in=__builtin_popcount(pos);
for(int k=0;;k++)
{
s=pos+pow(2,k)
if( __builtin_popcount(s) <= in )
{
in = __builtin_popcount(s)
pos = s;
if(pos>N)       break;
add M water ballons at bucket pos
}
}
pos = pos - N
}
pos = back+plus;
if(pos>N) pos-=N;
}
}
Jaggu is too lazy to put the water ballons in the bucket. Afraid that he might be caught for not doing what his sister told him to do so, he asks your help to provide correct answers for each of his sister's query. .

Input Format

First line contains Q, number of queries to follow.

Next Q line follows , which can be either an Update Query or Report Query.Each Update Query is followed by atleast 1 report query.

Output Format

For each report query , output the answer in a separate line.

Constraints

1 ≤ Q ≤ 2 * 105

1 ≤ pos1,pos2,pos ≤ 106

pos1 ≤ pos2

1 ≤ M ≤ 10

1 ≤ plus ≤ 999999```

### Solution :

```                            ```Solution in C :

In    C++  :

#include<iostream>
#include<string>
using namespace std;

long long tree[1000001];
int bitcount[2000001];

void init()
{
for (int i = 0; i <= 2000000; i++)
for (int j = 0; j < 32; j++)
if (i&(1 << j))
bitcount[i]++;
}

void update(int idx, int val)
{
while (idx <= 1000000)
{
tree[idx] += val;
idx += (idx & -idx);
}
}

{
long long sum = 0;
while (idx > 0)
{
sum += tree[idx];
idx -= (idx & -idx);
}
return sum;
}

void update(int pos, int M, int plus)
{
int N = 1000000;
for (int i = 1; i <= 50; i++)
{
int back = pos, j;
for (j = 1; j <= 1000; j++)
{
int s, in = bitcount[pos];
if (pos == 48576) break;
update(pos, M);
for (int k = 0;; k++)
{
s = pos + (k == 0 ? 1 : (1 << (k - 1)));
if (bitcount[s] <= in)
{
in = bitcount[s];
pos = s;
if (pos > N) break;
update(pos, M);
}
}
pos -= N;
}

update(48576, (1000 - j + 1) * M);
update(48640, (1000 - j + 1) * M);
update(49152, (1000 - j + 1) * M);
update(65536, (1000 - j + 1) * M);
update(131072, (1000 - j + 1) * M);
update(262144, (1000 - j + 1) * M);
update(524288, (1000 - j + 1) * M);

pos = back + plus;
if (pos > N) pos -= N;
}
}

int main()
{
int q;
cin >> q;
init();
while (q--)
{
char ch;
cin >> ch;
if (ch == 'U')
{
int pos, M, plus;
cin >> pos >> M >> plus;
update(pos, M, plus);
}
else
{
int pos1, pos2;
cin >> pos1 >> pos2;
long long sum1 = read(pos1 - 1);
cout << sum2 - sum1 << endl;
}
}
return 0;
}

In   Java  :

import java.io.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;

public class Solution {

public static void solve(Input in, PrintWriter out)
throws IOException {
int n = 1000000;
int[] next = new int[n];
int[] next2 = new int[n];
for (int i = n - 1; i >= 0; --i) {
next[i] = i | (i + 1);
if (next[i] >= n) {
next2[i] = next[i] - n;
} else {
next2[i] = next2[next[i]];
}
}
long[] f = new long[n];
int qs = in.nextInt();
for (int q = 0; q < qs; ++q) {
if (in.next().equals("U")) {
int pos = in.nextInt() - 1;
long m = in.nextLong();
for (int it = 0; it < 50; ++it) {
int x = (pos + it * add) % n;
int incs = 1000;
while (incs > 0) {
if (next2[x] == x) {
inc(x, m * incs, f);
incs = 0;
} else {
inc(x, m, f);
--incs;
x = next2[x];
}
}
}
} else {
int pos1 = in.nextInt() - 1;
int pos2 = in.nextInt() - 1;
out.println(get(pos2, f) - get(pos1 - 1, f));
}
}
}

private static long get(int x, long[] f) {
long r = 0;
while (x >= 0) {
r += f[x];
x = (x & (x + 1)) - 1;
}
return r;
}

private static void inc(int x, long m, long[] f) {
while (x < f.length) {
x = x | (x + 1);
}
}

public static void main(String[] args) throws IOException {
PrintWriter out = new PrintWriter(System.out);
out.close();
}

static class Input {
StringBuilder sb = new StringBuilder();

this.in = in;
}

public Input(String s) {
}

public String next() throws IOException {
sb.setLength(0);
while (true) {
if (c == -1) {
return null;
}
if (" \n\r\t".indexOf(c) == -1) {
sb.append((char)c);
break;
}
}
while (true) {
if (c == -1 || " \n\r\t".indexOf(c) != -1) {
break;
}
sb.append((char)c);
}
return sb.toString();
}

public int nextInt() throws IOException {
return Integer.parseInt(next());
}

public long nextLong() throws IOException {
return Long.parseLong(next());
}

public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
}
}

In    C  :

#include<stdio.h>

int i,j,k,n,m,pos,plus,pos1,pos2,q,t,back;
long long a[1111111];
char c;

int lowbit(int i) {
return i & -i;
}

void update(int index, int delta) {
for (int i = index; i <= 1000000; i += lowbit(i))
a[i] += delta;
}

long long sum(int index) {
long res = 0;
for (; index > 0; index -= lowbit(index))
res += a[index];
return res;
}

main()
{
scanf("%d",&q);
for (t=1; t<=q; t++)
{
scanf("%c",&c);
while (c!='R' && c!='U') scanf("%c",&c);
if (c=='R') scanf("%d%d",&pos1,&pos2); else scanf("%d%d%d",&pos,&m,&plus);
if (c=='R') printf("%lld\n",sum(pos2)-sum(pos1-1)); else
for (i=1; i<=50; i++)
{
back=pos;
while (pos<=1000000)
{
update(pos,m);
pos+=lowbit(pos);
}
pos-=1000000;
while (pos<=1000000)
{
update(pos,m);
pos+=lowbit(pos);
}
pos-=1000000;
while (pos<=1000000)
{
update(pos,998*m);
pos+=lowbit(pos);
}
pos=back+plus;
if (pos>1000000) pos-=1000000;
}
}
}```
```

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