Palindromic Subsets
Problem Statement :
Consider a lowercase English alphabetic letter character denoted by c. A shift operation on some c turns it into the next letter in the alphabet. For example, and ,shift(a) = b , shift(e) = f, shift(z) = a . Given a zero-indexed string, s, of n lowercase letters, perform q queries on s where each query takes one of the following two forms: 1 i j t: All letters in the inclusive range from i to j are shifted t times. 2 i j: Consider all indices in the inclusive range from i to j. Find the number of non-empty subsets of characters c1, c2, . . . , ck , where , i < index of c1 < index of c2 < . . . < index of ck < j ) such that characters c1, c2, . . . , ck , can be rearranged to form a palindrome. Then print this number modulo 10^9 + 7 on a new line. Two palindromic subsets are considered to be different if their component characters came from different indices in the original string. Note Two palindromic subsets are considered to be different if their component characters came from different indices in the original string. Input Format The first line contains two space-separated integers describing the respective values of n and q. The second line contains a string of n lowercase English alphabetic letters (i.e., a through z) denoting s. Each of the q subsequent lines describes a query in one of the two formats defined above. Output Format For each query of type 2 (i.e., 2 i j), print the number of non-empty subsets of characters satisfying the conditions given above, modulo 10 ^ 9 + 7, on a new line.
Solution :
Solution in C :
In C ++ :
#include <algorithm>
#include <cassert>
#include <cstring>
#include <iostream>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)
#define TRACE(x) cout << #x << " = " << x << endl
#define _ << " _ " <<
typedef long long llint;
const int MAX = 100100;
const int off = 1<<17;
const int mod = 1e9 + 7;
inline int add(int a, int b) {
return a+b >= mod ? a+b-mod : a+b;
}
inline int sub(int a, int b) {
return a >= b ? a-b : a-b+mod;
}
inline int mul(int a, int b) {
return llint(a)*b % mod;
}
vector<int> T[2*off];
int S[2*off];
void sh(int i, int x) {
S[i] = (S[i] + x) % 26;
rotate(T[i].begin(), T[i].begin() + ((26-x) % 26), T[i].end());
}
void propagate(int i) {
if (S[i]) {
sh(i*2, S[i]);
sh(i*2+1, S[i]);
S[i] = 0;
}
}
vector<int> merge(const vector<int>& a, const vector<int>& b) {
vector<int> c(26);
REP(i, 26) c[i] = a[i] + b[i];
return c;
}
void shift(int i, int lo, int hi, int a, int b, int k) {
if (lo >= b || hi <= a) return;
if (lo >= a && hi <= b) { sh(i, k); return; }
propagate(i);
shift(i*2, lo, (lo+hi)/2, a, b, k);
shift(i*2+1, (lo+hi)/2, hi, a, b, k);
T[i] = merge(T[i*2], T[i*2+1]);
}
vector<int> count(int i, int lo, int hi, int a, int b) {
if (lo >= b || hi <= a) return vector<int>(26, 0);
if (lo >= a && hi <= b) return T[i];
propagate(i);
return merge(count(i*2, lo, (lo+hi)/2, a, b), count(i*2+1, (lo+hi)/2, hi, a, b));
}
char s[MAX];
int pw[MAX];
int main(void) {
pw[0] = 1;
FOR(i, 1, MAX) pw[i] = mul(pw[i-1], 2);
int n, q;
scanf("%d %d", &n, &q);
scanf("%s", s);
REP(i, 2*off) T[i].resize(26);
REP(i, n) T[off+i][s[i]-'a']++;
for (int i = off-1; i >= 0; --i) REP(j, 26) T[i][j] = T[2*i][j] + T[2*i+1][j];
REP(i, q) {
int tip;
scanf("%d", &tip);
if (tip == 1) {
int a, b, t;
scanf("%d %d %d", &a, &b, &t);
t %= 26;
shift(1, 0, off, a, b+1, t);
}
if (tip == 2) {
int a, b;
scanf("%d %d", &a, &b);
vector<int> v = count(1, 0, off, a, b+1);
int ans = 0;
REP(j, 27) {
int ways = 1;
REP(k, 26)
if (v[k] && k != j) ways = mul(ways, pw[v[k]-1]);
if (j != 26) {
if (v[j] == 0) ways = 0;
else ways = mul(ways, pw[v[j]-1]);
}
ans = add(ans, ways);
}
ans = sub(ans, 1);
printf("%d\n", ans);
}
}
return 0;
}
In C :
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
#define M 1000000007
char s[100006];
int ret[32];
int d[1<<18][32];
int f[100006];
void init(int t,int l,int r) {
int i;
if (r-l==1) {
d[t][s[l]-'a']++;
return ;
}
init(t<<1,l,(l+r)/2);
init(t<<1|1,(l+r)/2,r);
for(i=0;i<26;i++) d[t][i]=d[t<<1][i]+d[t<<1|1][i];
}
void shift(int a[],int t) {
int b[32],i;
for(i=0;i<26;i++) b[(i+t)%26]=a[i];
for(i=0;i<26;i++) a[i]=b[i];
}
void update(int t,int l,int r,int a,int b,int val) {
if (l==a && r==b) {
d[t][26]=(d[t][26]+val)%26;
return ;
}
shift(d[t],d[t][26]);
d[t<<1][26]+=d[t][26];
d[t<<1|1][26]+=d[t][26];
d[t][26]=0;
if (b<=(l+r)/2) {
update(t<<1,l,(l+r)/2,a,b,val);
} else if (a>=(l+r)/2) {
update(t<<1|1,(l+r)/2,r,a,b,val);
} else {
update(t<<1,l,(l+r)/2,a,(l+r)/2,val);
update(t<<1|1,(l+r)/2,r,(l+r)/2,b,val);
}
int i;
for(i=0;i<26;i++) d[t][i]=0;
for(i=0;i<26;i++) d[t][(i+d[t<<1][26])%26]+=d[t<<1][i];
for(i=0;i<26;i++) d[t][(i+d[t<<1|1][26])%26]+=d[t<<1|1][i];
}
void query(int t,int l,int r,int a,int b) {
// printf("%d %d %d %d %d %d %d\n",t,l,r,d[t][0],d[t][1],d[t][2],d[t][26]);
if (l==a && r==b) {
int i;
for(i=0;i<26;i++) ret[(i+d[t][26])%26]+=d[t][i];
return ;
}
shift(d[t],d[t][26]);
d[t<<1][26]+=d[t][26];
d[t<<1|1][26]+=d[t][26];
d[t][26]=0;
if (b<=(l+r)/2) {
query(t<<1,l,(l+r)/2,a,b);
} else if (a>=(l+r)/2) {
query(t<<1|1,(l+r)/2,r,a,b);
} else {
query(t<<1,l,(l+r)/2,a,(l+r)/2);
query(t<<1|1,(l+r)/2,r,(l+r)/2,b);
}
}
int main(){
int n,m,i;
scanf("%d %d",&n,&m);
scanf("%s",s);
init(1,0,n);
for(f[0]=i=1;i<=n;i++) if ((f[i]=f[i-1]+f[i-1])>=M) f[i]-=M;
for(;m--;) {
int op,l,r,val;
scanf("%d %d %d",&op,&l,&r);
if (op==1) {
scanf("%d",&val);
update(1,0,n,l,r+1,val);
} else {
int d1=1,d2=0;
memset(ret,0,sizeof(ret));
query(1,0,n,l,r+1);
// printf("%d %d %d\n",ret[0],ret[1],ret[2]);
for(i=0;i<26;i++) {
if (!ret[i]) continue;
d2=(long long)(d2+d1)*f[ret[i]-1]%M;
d1=(long long)d1*f[ret[i]-1]%M;
}
if ((d1+=d2-1)>=M) d1-=M;
printf("%d\n",d1);
}
}
return 0;
}
In Python3 :
#!/bin/python3
import sys
n,q = input().strip().split(' ')
n,q = [int(n),int(q)]
s = input().strip()
sa = [ord(i) - ord('a') for i in s]
for a0 in range(q):
qu = input().strip().split(' ')
if int(qu[0]) == 1:
start = int(qu[1])
end =int(qu[2])
t = int(qu[3])
for i in range(start, end + 1):
sa[i] = (sa[i] + t) % 26
else:
start = int(qu[1])
end =int(qu[2])
d = [0] * 27
for i in range(start, end + 1):
d[sa[i]] += 1
ans = 1
for i in range(27):
if d[i]:
ans = (ans * pow(2, d[i] - 1) )% 1000000007
g = ans
ans = (ans - 1 )% 1000000007
for i in range(27):
if d[i]:
ans = (ans + g) % 1000000007
print(ans)
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class AliceBobSillyGames {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int g = in.nextInt();
for(int a0 = 0; a0 < g; a0++){
int n = in.nextInt();
// your code goes here
getSetArray(n);
}
}
static void getSetArray(int n) {
int [] numberSet = new int[n];
for(int i = 1; i <= n; i++) {
numberSet[i-1] = i;
}
int flag = 0;
for(int i = 2; i <= n; i++) {
if(isPrime(i)) {
/* System.out.print(i +" ");*/
int k = 1;
while(k <= n) {
if(i*k - 1 < n) {
numberSet[i*k - 1] = 0;
}
k++;
//k = k*i;
}
flag++;
}
}
if(flag % 2 == 0) {
System.out.println("Bob");
}
else {
System.out.println("Alice");
}
}
static public boolean isPrime(long num) {
if ( num < 2 ) return false;
for (int i = 2; i <= Math.sqrt(num); i++) {
if ( num % i == 0 ) {
return false;
}
}
return true;
}
}
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