Polynomial Division
Problem Statement :
Consider a sequence, c0, c1, . . . , cn-1 , and a polynomial of degree 1 defined as Q(x ) = a * x + b. You must perform q queries on the sequence, where each query is one of the following two types: 1 i x: Replace ci with x. 2 l r: Consider the polynomial and determine whether is divisible by over the field , where . In other words, check if there exists a polynomial with integer coefficients such that each coefficient of is divisible by Q( x ) = a*x + b. If a valid exists, print Yes on a new line; otherwise, print No. Given the values of , , , and queries, perform each query in order. Input Format The first line contains four space-separated integers describing the respective values of n (the length of the sequence), a (a coefficient in Q( x ) ), b (a coefficient in Q( x ) ), and q(the number of queries). The second line contains n space-separated integers describing c0, c1, . . . , cn-1. Each of the q subsequent lines contains three space-separated integers describing a query of either type 1 or type 2. Output Format For each query of type 2, print Yes on a new line if Q(x) is a divisor of P(x); otherwise, print No instead. Sample Input 0 3 2 2 3 1 2 3 2 0 2 1 2 1 2 0 2 Sample Output 0 No Yes
Solution :
Solution in C :
In C ++ :
#include <bits/stdc++.h>
using namespace std;
typedef pair<long long int,long long int> pll;
const int T = (1<<17);
const long long int MOD = 1000000007;
long long int X;
pll seg[2*T];
long long int power(long long int a, int b)
{
if(!b)
return 1;
long long int ans = power(a,b/2);
ans = (ans*ans)%MOD;
if(b%2)
ans = (ans*a)%MOD;
return ans;
}
pll seg_merge(pll &v1, pll &v2)
{
pll ret;
ret.first = (v1.first + v2.first*power(X,v1.second))%MOD;
ret.second = v1.second + v2.second;
return ret;
}
pll que(int root, int lm, int rm, int u, int v)
{
if(u <= lm && rm <= v)
return seg[root];
int mid = (lm + rm)/2;
if(u <= mid)
{
pll lval = que(2*root, lm, mid, u, v);
if(mid < v)
{
pll rval = que(2*root+1, mid+1, rm, u, v);
return seg_merge(lval,rval);
}
return lval;
}
pll rval = que(2*root+1, mid+1, rm, u, v);
return rval;
}
int main()
{
int n,a,b,q;
scanf("%d %d %d %d", &n, &a, &b, &q);
X = ((MOD - b)*power(a,MOD-2))%MOD;
for (int i = 0; i < n; ++i)
{
scanf("%lld", &seg[T+i].first);
seg[T+i].second = 1;
}
for (int i = T-1; i >= 1; --i)
seg[i] = seg_merge(seg[2*i],seg[2*i+1]);
while(q--)
{
int ch;
scanf("%d", &ch);
if(ch == 1)
{
int pos, val;
scanf("%d %d", &pos, &val);
pos+=T;
seg[pos].first = val;
pos/=2;
while(pos)
{
seg[pos] = seg_merge(seg[2*pos],seg[2*pos+1]);
pos/=2;
}
}
else
{
int l,r;
scanf("%d %d", &l, &r);
l+=T;
r+=T;
pll ans = que(1, T, 2*T-1, l, r);
if(ans.first)
printf("No\n");
else
printf("Yes\n");
}
}
return 0;
}
In Java :
import java.io.*;
import java.util.*;
public class D {
BufferedReader br;
PrintWriter out;
StringTokenizer st;
boolean eof;
static final int P = 1_000_000_007;
int pow(int a, int b) {
int ret = 1;
for (; b > 0; b >>= 1) {
if ((b & 1) == 1) {
ret = (int) ((long) ret * a % P);
}
a = (int) ((long) a * a % P);
}
return ret;
}
void add(long[] f, int pos, long delta) {
for (int i = pos; i < f.length; i |= i + 1) {
f[i] += delta;
}
}
long get(long[] f, int pos) {
long ret = 0;
for (int i = pos; i >= 0; i = (i & (i + 1)) - 1) {
ret += f[i];
}
return ret;
}
void solve() throws IOException {
int n = nextInt();
int a = nextInt();
int b = nextInt();
int q = nextInt();
int x;
if (a == 0) {
x = 1;
} else {
x = (int) ((long) b * pow(a, P - 2) % P);
if (x != 0) {
x = P - x;
}
}
int[] pow = new int[n];
pow[0] = 1;
for (int i = 1; i < n; i++) {
pow[i] = (int) ((long) pow[i - 1] * x % P);
}
int[] arr = new int[n];
int[] arrX = new int[n];
long[] fen = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
arrX[i] = (int) ((long) arr[i] * pow[i] % P);
add(fen, i, arrX[i]);
}
for (int i = 0; i < q; i++) {
int type = nextInt();
if (type == 1) {
int pos = nextInt();
int val = nextInt();
add(fen, pos, -arrX[pos]);
arr[pos] = val;
arrX[pos] = (int) ((long) val * pow[pos] % P);
add(fen, pos, arrX[pos]);
} else {
int l = nextInt();
int r = nextInt();
// [l; r]
long sumR = get(fen, r);
long sumL = get(fen, l - 1);
if (a == 0 && b == 0) {
out.println(sumR - sumL == 0 ? "Yes" : "No");
} else if (a == 0 && b != 0) {
out.println("Yes");
} else if (a != 0 && b == 0) {
out.println(arr[l] == 0 ? "Yes" : "No");
} else {
out.println((sumR - sumL) % P == 0 ? "Yes" : "No");
}
}
}
}
D() throws IOException {
br = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
solve();
out.close();
}
public static void main(String[] args) throws IOException {
new D();
}
String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
eof = true;
return null;
}
}
return st.nextToken();
}
String nextString() {
try {
return br.readLine();
} catch (IOException e) {
eof = true;
return null;
}
}
int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
}
In C :
#include <stdio.h>
#include <stdlib.h>
#define MOD 1000000007
void build(int v,int tl,int tr);
void update(int v,int tl,int tr,int pos,int new_val);
long long sum(int v,int tl,int tr,int l,int r);
long long modInverse(long long a,long long mod);
int min(int x,int y);
int max(int x,int y);
int c[100000];
long long dp[100000],t[400000];
int main(){
int n,a,b,q,x,y,a_inv,i;
scanf("%d%d%d%d",&n,&a,&b,&q);
a_inv=modInverse(a,MOD);
for(i=dp[0]=1;i<100000;i++)
dp[i]=(MOD-dp[i-1]*b%MOD*a_inv%MOD)%MOD;
for(i=0;i<n;i++)
scanf("%d",c+i);
build(1,0,n-1);
while(q--){
scanf("%d",&x);
if(x==1){
scanf("%d%d",&x,&y);
update(1,0,n-1,x,y);
}
else{
scanf("%d%d",&x,&y);
if(sum(1,0,n-1,x,y))
printf("No\n");
else
printf("Yes\n");
}
}
return 0;
}
void build(int v,int tl,int tr){
int tm;
if(tl==tr)
t[v]=c[tl];
else{
tm=(tl+tr)/2;
build(v*2,tl,tm);
build(v*2+1,tm+1,tr);
t[v]=(t[v*2]+t[v*2+1]*dp[tm-tl+1])%MOD;
}
return;
}
void update(int v,int tl,int tr,int pos,int new_val){
int tm;
if(tl==tr)
t[v]=new_val;
else{
tm=(tl+tr)/2;
if(pos<=tm)
update(v*2,tl,tm,pos,new_val);
else
update(v*2+1,tm+1,tr,pos,new_val);
t[v]=(t[v*2]+t[v*2+1]*dp[tm-tl+1])%MOD;
}
return;
}
long long sum(int v,int tl,int tr,int l,int r){
int tm,temp;
if(l>r)
return 0;
if(l==tl && r==tr)
return t[v];
tm=(tl+tr)/2;
if(min(r,tm)>=l)
temp=min(r,tm)-l+1;
else
temp=0;
return (sum(v*2,tl,tm,l,min(r,tm))+sum(v*2+1,tm+1,tr,max(l,tm+1),r)*dp[temp])%MOD;
}
long long modInverse(long long a,long long mod){
long long b0 = mod, t, q;
long long x0 = 0, x1 = 1;
while (a > 1) {
q = a / mod;
t = mod; mod = a % mod; a = t;
t = x0; x0 = x1 - q * x0; x1 = t;
}
if (x1 < 0) x1 += b0;
return x1;
}
int min(int x,int y){
return (x<y)?x:y;
}
int max(int x,int y){
return (x>y)?x:y;
}
In Python3 :
def init(R, x, p):
T = [R]
while len(R) > 1:
if len(R) & 1: R.append(0)
R = [(R[i] + x * R[i+1]) % p for i in range(0,len(R),2)]
x = (x * x) % p
T.append(R)
return T
def update(T, i, x, p):
S = T[0]
for j in range(1, len(T)):
R = T[j]
i >>= 1
R[i] = (S[2*i] + x*S[2*i+1]) % p
S = R
x = (x * x) % p
def query(T, i, x, p):
if i == 0: return T[-1][0]
s = 0
y = 1
for j in range(len(T)-1):
if i & 1:
s = (s + y * T[j][i]) % p
y = (y * x) % p
i = (i + 1) >> 1
x = (x * x) % p
return s
p = 10**9 + 7
n, a, b, q = map(int, input().split())
c = [int(x) for x in input().split()]
x = (-b * pow(a,p-2,p)) % p
T = init(c, x, p)
for Q in range(q):
k, a, b = map(int, input().split())
if k == 1:
c[a] = b
update(T, a, x, p)
elif k == 2:
y = (query(T,a,x,p) - query(T,b+1,x,p) * pow(x,b-a+1,p)) % p
print('No' if y else 'Yes')
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