Arithmetic Progressions


Problem Statement :


Let F(a,d) denote an arithmetic progression (AP) with first term  and common difference , i.e.  denotes an infinite . You are given  APs => . Let  denote the sequence obtained by multiplying these APs.

Multiplication of two sequences is defined as follows. Let the terms of the first sequence be , and terms of the second sequence be . The sequence obtained by multiplying these two sequences is
If  are the terms of a sequence, then the terms of the first difference of this sequence are given by  calculated as  respectively. Similarly, the second difference is given by , and so on.

We say that the  difference of a sequence is a constant if all the terms of the  difference are equal.

Let  be a sequence defined as => 
Similarly,  is defined as => product of .

Task:
Can you find the smallest  for which the  difference of the sequence  is a constant? You are also required to find this constant value.

You will be given many operations. Each operation is of one of the two forms:

1) 0 i j => 0 indicates a query . You are required to find the smallest  for which the  difference of  is a constant. You should also output this constant value.

2) 1 i j v => 1 indicates an update . For all , we update .

Input Format
The first line of input contains a single integer , denoting the number of APs.
Each of the next  lines consists of three integers  .
The next line consists of a single integer , denoting the number of operations. Each of the next  lines consist of one of the two operations mentioned above.  

Output Format
For each query, output a single line containing two space-separated integers  and .  is the smallest value for which the  difference of the required sequence is a constant.  is the value of this constant. Since  might be large, output the value of  modulo 1000003.

Note:  will always be such that it fits into a signed 64-bit integer. All indices for query and update are 1-based. Do not take modulo 1000003 for .


Solution :



title-img


                            Solution in C :

In    C++  :









#include <cstdio>
#include <cstring>
#include <string>
using namespace std;

const int N=100005;
const int M=1000003;
typedef long long i64;

struct node {
int ans;
int mul;
i64 sump;
i64 c;
};

node tree[N<<2];
int f[M];

int mul(i64 x,i64 y) {
return x*y%M;
}

int gao(int x,int y) {
if (x==0) {
return 0;
}
if (y==0) {
return 1;
}
if (y&1) {
y=gao(x,y-1);
return mul(y,x);
}
y=gao(x,y>>1);
return mul(y,y);
}

void build(int ind,int left,int right) {
int mid;
tree[ind].ans=1;  
tree[ind].mul=1; 
tree[ind].sump=0; 
tree[ind].c=0;
if (left==right) {
return;
}
mid=(left+right)>>1;
build(ind<<1,left,mid);
build((ind<<1)|1,mid+1,right);
}

void update(int ind,int left,int right,
int ll,int rr,int x) {  
int mid,lson,rson;
if ((left==ll) && (right==rr)) {
tree[ind].sump+=(rr-ll+1)*x;
tree[ind].ans=mul(tree[ind].ans,gao(tree[ind].mul,x));
tree[ind].c+=x; 
return;
}
mid=(left+right)>>1;
lson=ind<<1;
rson=lson|1;
if (tree[ind].c) {   
tree[lson].sump+=(mid-left+1)*tree[ind].c;
tree[rson].sump+=(right-mid)*tree[ind].c;
tree[lson].ans=mul(tree[lson].ans,gao(
    tree[lson].mul,tree[ind].c));
tree[rson].ans=mul(tree[rson].ans,gao(
    tree[rson].mul,tree[ind].c));
tree[lson].c+=tree[ind].c;
tree[rson].c+=tree[ind].c;
tree[ind].c=0;
}  
if (rr<=mid) {
update(lson,left,mid,ll,rr,x);
}
else if (ll>mid) {
update(rson,mid+1,right,ll,rr,x);
}
else {
update(lson,left,mid,ll,mid,x);
update(rson,mid+1,right,mid+1,rr,x);
}
tree[ind].ans=mul(tree[lson].ans,tree[rson].ans);
tree[ind].sump=tree[lson].sump+tree[rson].sump;
}

void insert(int ind,int left,int right,
int x,int p,int d,int dp) { 
int mid;
tree[ind].sump+=p;
tree[ind].mul=mul(tree[ind].mul,d);
tree[ind].ans=mul(tree[ind].ans,dp);
if (left==right) {
return;
}
mid=(left+right)>>1;
if (x<=mid) {
insert(ind<<1,left,mid,x,p,d,dp);
}
else {
insert((ind<<1)|1,mid+1,right,x,p,d,dp);
}
}

pair<i64,int> query(int ind,int left,
int right,int ll,int rr) {
int mid,lson,rson;
pair<i64,int> templ,tempr;
if ((left==ll) && (right==rr)) {
return make_pair(tree[ind].sump,tree[ind].ans);
}
mid=(left+right)>>1;
lson=ind<<1;
rson=lson|1;
if (tree[ind].c) {   
tree[lson].sump+=(mid-left+1)*tree[ind].c;
tree[rson].sump+=(right-mid)*tree[ind].c;
tree[lson].ans=mul(
tree[lson].ans,gao(
tree[lson].mul,tree[ind].c));
tree[rson].ans=mul(
tree[rson].ans,gao(tree[rson].mul,tree[ind].c));
tree[lson].c+=tree[ind].c;
tree[rson].c+=tree[ind].c;
tree[ind].c=0;
}  
if (rr<=mid) {
return query(lson,left,mid,ll,rr);
}
else if (ll>mid) {
return query(rson,mid+1,right,ll,rr);
}
else {
templ=query(lson,left,mid,ll,mid);
tempr=query(rson,mid+1,right,mid+1,rr);
return make_pair(templ.first+tempr.first,
mul(templ.second,tempr.second));
}


}

int main() {
pair<i64,int> ans;
int i,j,x,y,n;
for (i=f[0]=1;i<M;++i) {
f[i]=mul(f[i-1],i);
}
scanf("%d",&n);
build(1,1,n);
for (i=1;i<=n;++i) {
scanf("%d",&x);
scanf("%d%d",&x,&y);
insert(1,1,n,i,y,x,gao(x,y));
}
for (scanf("%d",&i);i;--i) {
scanf("%d%d%d",&j,&x,&y);
if (j==0) {
ans=query(1,1,n,x,y);
printf("%Ld %d\n",
ans.first,(ans.first>=M)?0:mul(f[ans.first],ans.second));
}
else {
scanf("%d",&j);
update(1,1,n,x,y,j);
}
}
return 0;
}









In      Java  :







import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.Scanner;

public class Solution {
private static class SumTree {
private final int n;
private final long[] elements;
private final long[] operations;
private final long[] source;

public SumTree(long[] elements) {
n = elements.length;
this.elements = new long[4 * n + 1];
this.operations = new long[4 * n + 1];
this.source = new long[n];

initFromArray(elements, 0, n, 0);
}

protected long monoidSum(long arg1, long arg2) {
return arg1 + arg2;
}

protected long operationSum(long arg1, 
long arg2) {
return arg1 + arg2;
}

protected long applyOperation(long monoidSum, 
int l, int r, long operation) {
return monoidSum + (r - l) * operation;
}

// all [l,r)
private void initFromArray(long[] elements,
 int l, int r, int num) {
if (r == l + 1) {
this.elements[num] = elements[l];
this.source[l] = elements[l];
} else {
int m = (l + r) / 2;
initFromArray(elements, l, m, num * 2 + 1);
initFromArray(elements, m, r, num * 2 + 2);
this.elements[num] = monoidSum(this.elements[num * 2 + 1],
 this.elements[num * 2 + 2]);
}
}

private long getTransformedSum(int l, int r, 
int ll, int rr, int num) {
if ((l == ll) && (r == rr)) {
return elements[num];
} else {
int m = (ll + rr) / 2;
long interimResult;
if (r <= m) {
interimResult = getTransformedSum(l, r, ll,
 m, num * 2 + 1);
} else if (l >= m) {
interimResult = getTransformedSum(l, r, m, 
rr, num * 2 + 2);
} else {
interimResult = monoidSum(
getTransformedSum(l, m, ll, m, num * 2 + 1),
getTransformedSum(m, r, m, rr, num * 2 + 2)
);
}
return applyOperation(interimResult, l, r, 
operations[num]);
}
}

public long getTransformedSum(int l, int r) {
return getTransformedSum(l, r, 0, n, 0);
}

private void applyOperation(int l, int r, int ll,
 int rr, int num, long operation) {
if ((l == ll) && (r == rr)) {
operations[num] = operationSum(operations[num], operation);
} else {
int m = (ll + rr) / 2;
if (l < m) {
applyOperation(l, Math.min(m, r), ll, m, 
num * 2 + 1, operation);
}
if (m < r) {
applyOperation(Math.max(m, l), r, m, rr, 
num * 2 + 2, operation);
}
}
if (rr == ll + 1) {
elements[num] = applyOperation(source[l], ll,
 rr, operations[num]);
} else {
elements[num] = applyOperation(monoidSum(
    elements[num * 2 + 1], elements[num * 2 + 2]), 
    ll, rr, operations[num]);
}
}

public void applyOperation(int l, int r, 
long operation) {
applyOperation(l, r, 0, n, 0, operation);
}
}

private static class MultTree {
private static final long[] INVERSE = 
new long[(int) MODULO];

static {
int v = 1;
int modulo = (int) MODULO;
int[] powers = new int[modulo];
for (int i = 0; i < modulo; i++) {
powers[i] = v;
v = (v * 2) % modulo;
}
for (int i = 0; i < modulo; i++) {
INVERSE[powers[i]] = powers[modulo - 1 - i];
}
}

private final int n;
private final int[] elements;

public MultTree(long[] elements) {
n = elements.length;
this.elements = new int[n];

long current = 1;
for (int i = 0; i < elements.length; i++) {
current = (current * elements[i]) % MODULO;
this.elements[i] = (int) current;
}
}

// all [l,r)
public long getTransformedSum(int l, int r) {
if (l == 0) {
return elements[r - 1];
} else {
int rV = elements[r - 1];
int lV = elements[l - 1];
// should return rV-lV
return (1L * rV * INVERSE[lV]) % MODULO;
}
}
}

private static class MainTree {
private final int n;
private final long[] elements;
private final long[] operations;
private final long[] source;

private final MultTree dTree;

// all [l,r)
private void initFromArray(long[] elements,
 int l, int r, int num) {
if (r == l + 1) {
this.elements[num] = elements[l];
this.source[l] = elements[l];
} else {
int m = (l + r) / 2;
initFromArray(elements, l, m, num * 2 + 1);
initFromArray(elements, m, r, num * 2 + 2);
this.elements[num] = monoidSum(this.elements[num * 2 + 1], 
this.elements[num * 2 + 2]);
}
}

private long getTransformedSum(int l, int r, 
int ll, int rr, int num) {
if ((l == ll) && (r == rr)) {
return elements[num];
} else {
int m = (ll + rr) / 2;
long interimResult;
if (r <= m) {
interimResult = getTransformedSum(l, r, ll, 
m, num * 2 + 1);
} else if (l >= m) {
interimResult = getTransformedSum(l, r, m,
 rr, num * 2 + 2);
} else {
interimResult = monoidSum(
getTransformedSum(l, m, ll, m, num * 2 + 1),
getTransformedSum(m, r, m, rr, num * 2 + 2)
);
}
return applyOperation(interimResult, l, r, 
operations[num]);
}
}

public long getTransformedSum(int l, int r) {
return getTransformedSum(l, r, 0, n, 0);
}

private void applyOperation(int l, int r,
 int ll, int rr, int num, long operation) {
if ((l == ll) && (r == rr)) {
operations[num] = operationSum(operations[num], 
operation);
} else {
int m = (ll + rr) / 2;
if (l < m) {
applyOperation(l, Math.min(m, r), ll, m,
 num * 2 + 1, operation);
}
if (m < r) {
applyOperation(Math.max(m, l), r, m, rr,
 num * 2 + 2, operation);
}
}
if (rr == ll + 1) {
elements[num] = applyOperation(source[l], ll, 
rr, operations[num]);
} else {
elements[num] = applyOperation(monoidSum(
    elements[num * 2 + 1], elements[num * 2 + 2]),
    ll, rr, operations[num]);
}
}

public void applyOperation(int l, int r, long operation) {
applyOperation(l, r, 0, n, 0, operation);
}

public MainTree(long[] elements, MultTree dTree) {
n = elements.length;
this.elements = new long[4 * n + 1];
this.operations = new long[4 * n + 1];
this.source = new long[n];

initFromArray(elements, 0, n, 0);
this.dTree = dTree;
}

protected long monoidSum(long arg1, long arg2) {
return (arg1 * arg2) % MODULO;
}

protected long operationSum(long arg1, long arg2) {
return arg1 + arg2;
}

protected long applyOperation(long monoidSum, 
int l, int r, long operation) {
return (monoidSum * pow(
dTree.getTransformedSum(l, r), operation)) % MODULO;
}
}

private static final long MODULO = 1000003;


private static final long[] FACTORIALS = 
new long[(int) MODULO + 1];

static {
FACTORIALS[0] = 1;
for (int n = 1; n <= MODULO; n++) {
FACTORIALS[n] = (FACTORIALS[n - 1] * n) % MODULO;
}
}

public static long factorial(long n) {
if (n > MODULO) {
return 0;
} else {
return FACTORIALS[(int) n];
}
}

public static long pow(long v, long p) {
if (p == 0) {
return 1;
} else {
long tmp = pow(v, p / 2);
tmp = (tmp * tmp) % MODULO;
return p % 2 == 0 ? tmp : (tmp * v) % MODULO;
}
}

private void run(Scanner input, PrintWriter output) {
int n = input.nextInt();
int[] a = new int[n];
long[] p = new long[n], d = new long[n], 
dp = new long[n];
for (int i = 0; i < n; i++) {
a[i] = input.nextInt();
d[i] = input.nextInt();
p[i] = input.nextInt();

dp[i] = pow(d[i], p[i]);
}

SumTree pTree = new SumTree(p);
MultTree dTree = new MultTree(d);
MainTree cTree = new MainTree(dp, dTree);

int q = input.nextInt();
for (int qNum = 0; qNum < q; qNum++) {
int type = input.nextInt();
int i = input.nextInt() - 1;
int j = input.nextInt() - 1;

if (type == 0) {
long pRes = pTree.getTransformedSum(i, j + 1);
long cRes;

if (pRes > MODULO) {
cRes = 0;
} else {
cRes = cTree.getTransformedSum(i, j + 1);
}

output.println(pRes + " " + (
    (cRes * factorial(pRes)) % MODULO));
} else {
int v = input.nextInt();

pTree.applyOperation(i, j + 1, (long) v);
cTree.applyOperation(i, j + 1, (long) v); 
}
}
}

public static void main(String[] args) throws IOException {
Scanner input = new Scanner(new BufferedInputStream(System.in));
PrintWriter output = new PrintWriter(System.out);

(new Solution()).run(input, output);

output.flush();
}
}











In   C  :







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

long long power(int d,int n)
{
	long long i,temp;
	i=1;
	temp=d;
	while(n>0)
	{
		if(n%2==1) i*=temp;
		temp*=temp;
		if(temp>1000003) temp%=1000003;
		if(i>1000003) i%=1000003;
		n=n/2;
	}
	return i;
}

int main()
{
	int i,j,k,v;
	int s,e;
	int n,q,qr;
	long long temp,temp1;
	int *a,*d,*p;
        long long *nfac;
	nfac=(long long *)malloc(1000003*sizeof(long long));
	nfac[0]=1;
	for(i=1;i<1000003;i++)
	{
		nfac[i]=nfac[i-1]*i;
		if(nfac[i]>1000003) nfac[i]%=1000003;
	}
	scanf("%d",&n);
	a=(int *)malloc(n*sizeof(int));
	d=(int *)malloc(n*sizeof(int));
	p=(int *)malloc(n*sizeof(int));
	for(i=0;i<n;i++) scanf("%d %d %d",&a[i],&d[i],&p[i]);
	scanf("%d",&q);
	for(k=0;k<q;k++)
	{
		scanf("%d",&qr);
		if(qr==0)
		{
			temp=0;
			scanf("%d %d",&s,&e);
			for(i=s;i<=e;i++) temp+=p[i-1];
			printf("%d ",temp);
			if(temp>=1000003) 
			{
				printf("%d\n",0);
			}
			else
			{
				temp1=nfac[temp];
				for(i=s;i<=e;i++)
				{
					temp1*=power(d[i-1],p[i-1]);
					if(temp1>1000003) temp1%=1000003;
				}
				printf("%d\n",temp1);
			}
		}
		else
		{
			scanf("%d %d %d",&s,&e,&v);
			for(i=s;i<=e;i++) p[i-1]+=v;
		}
	}
}









In   Python3  :









from array import array

f = array('i', [1])

def mod_factorial(n, r):
    if n < len(f):
        return f[n]
    res = f[-1]
    for i in range(len(f), n+1):
        res = (res*i) % r
        f.append(res)
    return f[-1]

r = 1000003

n = int(input())

d = array('i')
p = array('i')
dp = array('i', [-1])*n
for _ in range(n):
    a, b, c = map(int, input().split())
    d.append(b)
    p.append(c)

q = int(input())

for _ in range(q):
    a = input().split()
    if a[0] == '0':
        i, j = map(int, a[1:])
        k = sum(p[i-1:j])
        v = mod_factorial(k, r)
        for m in range(i-1, j):
            if dp[m] < 0:
                dp[m] = pow(d[m], p[m], r)
            v = (v * dp[m]) % r
                
        print(k, v)
    else:
        i, j, v = map(int, a[1:])
        for k in range(i-1, j):
            p[k] += v
            dp[k] = -1
                        




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