Hamming Distance
Problem Statement :
You are given a string , consisting of small latin letters 'a' and 'b'. You are also given queries to process. The queries are as follows: C : all the symbols in the string, starting at the , ending at the become equal to ; S : swap two consecutive fragments of the string, where the first is denoted by a substring starting from ending at and the second is denoted by a substring starting at ending at ; R : reverse the fragment of the string that starts at the symbol and ends at the one; W : output the substring of the string that starts at the symbol and ends at the one; H : output the Hamming distance between the consecutive substrings that starts at and respectively and have the length of . Everything is 1-indexed here. Input Format The first line of input contains a single integer the length of the string. The second line contains the initial string itself. The third line of input contains a single integer the number of queries. Then, there are lines, each denotes a query of one of the types above. Output Format For each query of the type W or the type H output an answer on the separate line of output.
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
void HandleC(char *S, int N) {
int l, r, ch;
scanf ("%d %d %c\n", &l, &r, (char *)&ch);
l--;
r--;
while (l <= r) {
S[l] = ch;
l++;
}
}
char tS[500001];
void HandleS(char *S, int N) {
int l1, l2, r1, r2, len1, len2, end=0;
int i, l, r;
scanf ("%d %d %d %d\n", &l1, &r1, &l2, &r2);
l1--;
l2--;
r1--;
r2--;
int indices[4][2] = {
{l2, r2+1}, {r1+1,l2}, {l1, r1+1}, {r2+1, N+1}
};
strncpy(tS+l1, S+l1, N-l1);
end = l1;
for (int i = 0; i < 4; i++){
int l = indices[i][0], r = indices[i][1];
if (l < r) {
strncpy(S+end, tS+l, r-l+1);
}
end += r - l;
}
/*
len1 = r1 - l1 + 1;
len2 = r2 - l2 + 1;
memcpy(tS, &S[l1], l2-l1);
memcpy(&S[l1], &S[l2], len2);
memcpy(&S[l1+len2], &tS[len1-1], (l2-l1)-len1);
memcpy(&S[l1+len2+l2-l1-len1], tS, len1);
*/
}
void HandleR(char *S, int N) {
int l, r, temp;
scanf ("%d %d\n", &l, &r);
l--;
r--;
while (l <= r) {
temp = S[l];
S[l] = S[r];
S[r] = temp;
l++;
r--;
}
}
void HandleW(char *S, int N) {
int l, r;
scanf ("%d %d\n", &l, &r);
l--;
r--;
while (l <= r) {
printf ("%c", S[l]);
l++;
}
printf ("\n");
}
void HandleH(char *S, int N) {
int l1, l2, len, hd = 0;
scanf ("%d %d %d\n", &l1, &l2, &len);
l1--;l2--;
while (len) {
if (S[l1] != S[l2])
hd++;
len--;
l1++;
l2++;
}
printf ("%d\n", hd);
}
int main() {
int N, M;
int ch;
char S[50001];
scanf ("%d\n%s\n%d\n", &N, S, &M);
while (M) {
scanf("%c ", (char *)&ch);
switch (ch) {
case 'C':
HandleC(S, N);
break;
case 'S':
HandleS(S, N);
break;
case 'R':
HandleR(S, N);
break;
case 'W':
HandleW(S, N);
break;
case 'H':
HandleH(S, N);
break;
default:
break;
}
M--;
}
return 0;
}
Solution in C++ :
In C ++ :
#include <bits/stdc++.h>
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define REP(I, N) for (int I = 0; I < (N); ++I)
#define REPP(I, A, B) for (int I = (A); I < (B); ++I)
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define RS(X) scanf("%s", (X))
#define CASET int ___T, case_n = 1; scanf("%d ", &___T); while (___T-- > 0)
#define MP make_pair
#define PB push_back
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
typedef long long LL;
using namespace std;
const int LEN = 20;
const int BOUND = 1<<LEN;
int rev[BOUND],bit_cnt[BOUND];
int ONES;
void init(){
REP(i,LEN)ONES|=1<<i;
REP(i,BOUND){
bit_cnt[i]=bit_cnt[i>>1]+(i&1);
int tmp=i;
REP(j,LEN){
rev[i]<<=1;
rev[i]|=tmp&1;
tmp>>=1;
}
}
}
int a[50010];
char s[50010];
int tmp1[50010],tmp2[50010],tmp3[50010];
void get(int tmp[],int st,int len){
if(!len){
tmp[0]=0;
return;
}
int w=len/LEN;
int r=st%LEN;
int x=st/LEN;
int mask1=(1<<r)-1;
int mask2=(1<<LEN)-1-mask1;
REP(i,w){
tmp[i]=((a[x+i]&mask2)>>r)|((a[x+i+1]&mask1)<<(LEN-r));
}
r=len-LEN*w;
tmp[w]=0;
REP(i,r){
if((a[(st+w*LEN+i)/LEN]>>((st+w*LEN+i)%LEN))&1)
tmp[w]|=1<<i;
}
}
void put(int tmp[],int st,int len){
if(!len)return;
int w=len/LEN;
int r=st%LEN;
int x=st/LEN;
int mask1=(1<<r)-1;
int mask2=(1<<LEN)-1-mask1;
int mask3=(1<<(LEN-r))-1;
int mask4=(1<<LEN)-1-mask3;
REP(i,w){
a[x+i]&=mask1;
a[x+i]|=(tmp[i]&mask3)<<r;
a[x+i+1]&=mask2;
a[x+i+1]|=(tmp[i]&mask4)>>(LEN-r);
}
r=len-LEN*w;
int xx=(st+w*LEN)/LEN;
int yy=(st+w*LEN)%LEN;
REP(i,r){
if(((tmp[w]>>i)&1) != ((a[xx]>>yy)&1))
a[xx]^=1<<yy;
yy++;
if(yy==LEN){
xx++;
yy=0;
}
}
}
int main(){
init();
DRII(N,M);
RS(s);
int sn=LEN(s);
{
int x=0,y=0;
REP(i,sn){
if(s[i]=='b')a[x]|=1<<y;
y++;
if(y==LEN){
x++;
y=0;
}
}
}
DRI(Q);
while(Q--){
//REP(i,5)printf("%d",a[i]);
//puts("");
char c[4];
RS(c);
if(c[0]=='C'){
DRII(ll,rr);
ll--;rr--;
int st=(ll+LEN-1)/LEN,ed=(rr+1)/LEN;
RS(c);
int len=(rr-ll+LEN)/LEN;
if(c[0]=='a'){
REP(i,len)tmp1[i]=0;
}
else{
REP(i,len)tmp1[i]=ONES;
}
put(tmp1,ll,rr-ll+1);
}
else if(c[0]=='S'){
DRII(ll1,rr1);
DRII(ll2,rr2);
ll1--;rr1--;ll2--;rr2--;
get(tmp1,ll1,rr1-ll1+1);
get(tmp2,rr1+1,ll2-1-rr1);
get(tmp3,ll2,rr2-ll2+1);
// printf("[[%d%d]]\n",tmp1[0],tmp1[1]);
//printf("[[%d]]\n",tmp3[0]);
put(tmp3,ll1,rr2-ll2+1);
put(tmp2,ll1+rr2-ll2+1,ll2-1-rr1);
put(tmp1,ll1+rr2-rr1,rr1-ll1+1);
}
else if(c[0]=='R'){
DRII(ll,rr);
ll--;rr--;
int len=rr-ll+1;
get(tmp1,ll,len);
//REP(i,5)printf("[%d]",tmp1[i]);
if(len%LEN==0){
int w=len/LEN;
REP(i,w)tmp2[i]=rev[tmp1[w-i-1]];
}
else{
int w=len/LEN;
int r=len%LEN;
int mask1=(1<<r)-1;
int mask2=(1<<LEN)-1-mask1;
REP(i,w)tmp2[i]=rev[((tmp1[w-i]&mask1)<<(LEN-r))|((tmp1[w-i-1]&mask2)>>r)];
tmp2[w]=0;
REP(i,r)
if((tmp1[0]>>(r-i-1))&1)tmp2[w]|=1<<i;
}
put(tmp2,ll,len);
}
else if(c[0]=='W'){
DRII(ll,rr);
ll--;
int x=ll/LEN;
int y=ll%LEN;
REPP(i,ll,rr){
int now=(a[x]>>y)&1;
if(now)putchar('b');
else putchar('a');
y++;
if(y==LEN){
x++;
y=0;
}
}
puts("");
}
else if(c[0]=='H'){
DRIII(st1,st2,len);
st1--;st2--;
get(tmp1,st1,len);
get(tmp2,st2,len);
len=(len+LEN-1)/LEN;
int an=0;
REP(i,len)an+=bit_cnt[tmp1[i]^tmp2[i]];
printf("%d\n",an);
}
}
return 0;
}
Solution in Java :
In Java :
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.Arrays;
public class Hamming {
public static void main(String[] args) throws IOException, InterruptedException {
Hamming solution = new Hamming(System.in);
// Thread.sleep(10000);
solution.run();
}
private class Scanner {
public final byte[] bytes;
public final int length;
public int pos = 0;
private Scanner(InputStream in) throws IOException {
bytes = new byte[2026000];
int read;
int pos = 0;
while((read = in.read(bytes, pos, bytes.length - pos)) != -1)
pos += read;
this.length = pos;
}
private int nextInt() {
int i = 0;
int d;
while((d = (bytes[pos++] - '0')) >= 0 && d <= 9) {
i = i*10 + d;
}
while(true) {
switch(bytes[pos]) {
case '\r':
case '\n':
case ' ':
pos++;
continue;
default:
// System.out.println("nextInt: " + i);
return i;
}
}
}
private String nextLine() {
int start = pos;
while(bytes[pos] != '\r' && bytes[pos] != '\n')
pos++;
int len = pos - start;
skipNewline();
// System.out.println("nextLine pos " + pos + ": " + new String(bytes, start, len));
return new String(bytes, start, len);
}
private void skipNewline() {
while(true) {
switch(bytes[pos]) {
case '\r':
case '\n':
pos++;
continue;
default:
return;
}
}
}
}
private final Scanner scan;
private final int N;
private final int M;
private final CharArraySolution solution;
private Hamming(InputStream in) throws IOException {
scan = new Scanner(in);
N = scan.nextInt();
solution = new CharArraySolution(scan.nextLine());
M = scan.nextInt();
}
private void run() {
for(int m=0; m<M; m++) {
switch (scan.bytes[scan.pos++]) {
case 'C':
scan.pos++;
solution.set(scan.nextInt(), scan.nextInt(), scan.bytes[scan.pos++]);
scan.skipNewline();
break;
case 'S':
scan.pos++;
solution.swap(scan.nextInt(), scan.nextInt(), scan.nextInt(), scan.nextInt());
break;
case 'R':
scan.pos++;
solution.reverse(scan.nextInt(), scan.nextInt());
break;
case 'W':
scan.pos++;
solution.write(scan.nextInt(), scan.nextInt());
break;
case 'H':
scan.pos++;
solution.hamming(scan.nextInt(), scan.nextInt(), scan.nextInt());
break;
}
}
System.out.print(solution.toString());
}
private class CharArraySolution {
private final int N;
private final char[] forward;
private final char[] reverse;
private final char[] as;
private final char[] bs;
private final char[] buf;
private final char[] out;
private int outpos = 0;
private CharArraySolution(String string) {
char[] chars = string.toCharArray();
N = chars.length;
forward = new char[N+1];
System.arraycopy(chars, 0, forward, 1, N);
reverse = new char[N+1];
for(int i=1; i<=N; i++) {
reverse[i] = forward[N-i+1];
}
as = new char[N+1];
Arrays.fill(as, 'a');
bs = new char[N+1];
Arrays.fill(bs, 'b');
// write(1,N);
buf = new char[N+1];
out = new char[2000000];
}
private void set(int l, int r, byte c) {
// System.out.println("set " + l + " " + r + " " + c);
int len = r-l+1;
if(c == 'a') {
System.arraycopy(as, 0, forward, l, len);
l = N-r+1;
System.arraycopy(as, 0, reverse, l, len);
} else {
System.arraycopy(bs, 0, forward, l, len);
l = N-r+1;
System.arraycopy(bs, 0, reverse, l, len);
}
// write(1,N);
}
private void swap(int l1, int r1, int l2, int r2) {
// System.out.println("swap " + l1 + " " + r1 + " " + l2 + " " + r2);
int len1 = r1-l1+1;
int len2 = r2-l2+1;
int lenm = l2-r1-1;
if(len1 > len2) {
System.arraycopy(forward, l1, buf, 0, len1);
System.arraycopy(forward, l2, forward, l1, len2);
System.arraycopy(forward, r1+1, forward, l1+len2, lenm);
System.arraycopy(buf, 0, forward, l1+len2+lenm, len1);
int oldl1 = l1;
int oldr1 = r1;
l1 = N-r2+1;
r1 = N-l2+1;
l2 = N-oldr1+1;
r2 = N-oldl1+1;
int oldlen1 = len1;
len1 = len2;
len2 = oldlen1;
System.arraycopy(reverse, l2, buf, 0, len2);
System.arraycopy(reverse, l1, reverse, r2-len1+1, len1);
System.arraycopy(reverse, r1+1, reverse, r2-len1-lenm+1, lenm);
System.arraycopy(buf, 0, reverse, l1, len2);
} else {
System.arraycopy(forward, l2, buf, 0, len2);
System.arraycopy(forward, l1, forward, r2-len1+1, len1);
System.arraycopy(forward, r1+1, forward, r2-len1-lenm+1, lenm);
System.arraycopy(buf, 0, forward, l1, len2);
int oldl1 = l1;
int oldr1 = r1;
l1 = N-r2+1;
r1 = N-l2+1;
l2 = N-oldr1+1;
r2 = N-oldl1+1;
int oldlen1 = len1;
len1 = len2;
len2 = oldlen1;
System.arraycopy(reverse, l1, buf, 0, len1);
System.arraycopy(reverse, l2, reverse, l1, len2);
System.arraycopy(reverse, r1+1, reverse, l1+len2, lenm);
System.arraycopy(buf, 0, reverse, l1+len2+lenm, len1);
}
// write(1,N);
}
private void reverse(int l, int r) {
// System.out.println("reverse " + l + " " + r);
int len = r-l+1;
System.arraycopy(forward, l, buf, 0, len);
System.arraycopy(reverse, N-r+1, forward, l, len);
System.arraycopy(buf, 0, reverse, N-r+1, len);
// write(1,N);
}
private void write(int l, int r) {
int len = r-l+1;
System.arraycopy(forward, l, out, outpos, len);
outpos += len;
out[outpos++] = '\n';
// System.out.println(Arrays.copyOfRange(reverse, N-r+1, N-l+2));
}
private void hamming(int p1, int p2, int l) {
// System.out.println("ham " + p1 + " " + p2 + " " + l);
int c = 0;
for(int i=0; i<l; i++) {
if(forward[p1++] != forward[p2++])
c++;
}
int bufpos = 10;
do {
buf[bufpos--] = (char)('0' + (c%10));
c /= 10;
} while(c>0);
System.arraycopy(buf, bufpos+1, out, outpos, 10 - bufpos);
outpos += 10 - bufpos;
out[outpos++] = '\n';
}
public String toString() {
return new String(out, 0, outpos);
}
}
}
Solution in Python :
In Python3 :
L = int(input())
S = input().strip()
M = int(input())
x = 0
z = 1
for c in S:
if c == 'b':
x += z
z *= 2
x += z
def prnt(n):
i = 1
s = ''
while i < z:
s += 'ab'[i & n > 0]
i *= 2
print(s)
#prnt(x)
for i in range(M):
e = input().strip()
#e = 'S 1 2 3 4'
q = e.split()
a = q[0]
if a in 'CR':
i, j = map(lambda c: int(c)-1, q[1:3])
mask1 = (1 << i) - 1
mask2 = (1 << (j+1)) - 1
mask3 = (z - 1) ^ mask2
mask2 ^= mask1
y = x & (mask1 | mask3)
if a == 'C' and q[3] == 'b':
y |= mask2
elif a == 'R':
m = (x & mask2) >> i
m |= (mask2 >> i) + 1
rev = int(bin(m)[2:][::-1], base=2) >> 1
y |= rev << i
x = y
#prnt(x)
elif a == 'S':
i, j, t, u = map(lambda c: int(c)-1, q[1:])
mask5 = z - 1
mask4 = (1 << (u+1)) - 1
mask5 ^= mask4
mask3 = (1 << t) - 1
mask4 ^= mask3
mask2 = (1 << (j+1)) - 1
mask3 ^= mask2
mask1 = (1 << i) - 1
mask2 ^= mask1
y = x & mask1
y |= (x & mask4) >> (t - i)
#mask4 >>= t - i
p = (x & mask3)
p |= (x & mask2) << (t - i)
k = u + 1 - (t-i)
if k <= j:
p >>= j - k + 1
else:
p <<= k - j - 1
y |= p
y |= x & mask5
x = y
#prnt(x)
#break
elif a == 'W':
i, j = map(lambda c: int(c)-1, q[1:])
k = 1 << i
m = 1 << j
r = ''
while k <= m:
r += 'ab'[k & x > 0]
k <<= 1
print(r)
elif a == 'H':
#prnt(x)
i, j, L = map(int, q[1:])
i -= 1
j -= 1
msk = [0, 0]
for t, u in enumerate([i, j]):
msk[t] = (1 << (u + L)) - 1
msk[t] ^= (1 << u) - 1
t = (x & msk[0]) >> i
u = (x & msk[1]) >> j
print(bin(t^u).count('1'))
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