**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 read(int 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);
long long sum2 = read(pos2);
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();
int add = in.nextInt();
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) {
long add = m;
while (x < f.length) {
f[x] += add;
add += m;
x = x | (x + 1);
}
}
public static void main(String[] args) throws IOException {
PrintWriter out = new PrintWriter(System.out);
solve(new Input(new BufferedReader(
new InputStreamReader(System.in))), out);
out.close();
}
static class Input {
BufferedReader in;
StringBuilder sb = new StringBuilder();
public Input(BufferedReader in) {
this.in = in;
}
public Input(String s) {
this.in = new BufferedReader(new StringReader(s));
}
public String next() throws IOException {
sb.setLength(0);
while (true) {
int c = in.read();
if (c == -1) {
return null;
}
if (" \n\r\t".indexOf(c) == -1) {
sb.append((char)c);
break;
}
}
while (true) {
int c = in.read();
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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