Maximizing the Function
Problem Statement :
Consider an array of binary integers (i.e., 's and 's) defined as . Let be the bitwise XOR of all elements in the inclusive range between index and index in array . In other words, . Next, we'll define another function, : Given array and independent queries, perform each query on and print the result on a new line. A query consists of three integers, , , and , and you must find the maximum possible you can get by changing at most elements in the array from to or from to . Note: Each query is independent and considered separately from all other queries, so changes made in one query have no effect on the other queries. Input Format The first line contains two space-separated integers denoting the respective values of (the number of elements in array ) and (the number of queries). The second line contains space-separated integers where element corresponds to array element . Each line of the subsequent lines contains space-separated integers, , and respectively, describing query . Output Format Print lines where line contains the answer to query (i.e., the maximum value of if no more than bits are changed).
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <stdlib.h>
int a[500000],odd_sum[500000],even_sum[500000];
int main(){
int n,q,x,y,k,odd,even,i;
scanf("%d%d",&n,&q);
for(i=0;i<n;i++){
scanf("%d",a+i);
if(i){
a[i]^=a[i-1];
if(a[i]){
odd_sum[i]=odd_sum[i-1]+1;
even_sum[i]=even_sum[i-1];
}
else{
odd_sum[i]=odd_sum[i-1];
even_sum[i]=even_sum[i-1]+1;
}
}
else
if(a[i]){
odd_sum[i]=1;
even_sum[i]=0;
}
else{
odd_sum[i]=0;
even_sum[i]=1;
}
}
while(q--){
scanf("%d%d%d",&x,&y,&k);
if(k)
printf("%lld\n",(y-x+2)/2*((long long)(y-x+3)/2));
else{
odd=odd_sum[y];
even=even_sum[y];
if(x){
odd-=odd_sum[x-1];
even-=even_sum[x-1];
}
if(!x || !a[x-1])
printf("%lld\n",odd*(long long)(even+1));
else
printf("%lld\n",even*(long long)(odd+1));
}
}
return 0;
}
Solution in C++ :
In C++ :
#include "bits/stdc++.h"
using namespace std;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
static const int INF = 0x3f3f3f3f; static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
typedef vector<int> vi; typedef pair<int, int> pii; typedef vector<pair<int, int> > vpii; typedef long long ll;
template<typename T, typename U> static void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> static void amax(T &x, U y) { if(x < y) x = y; }
int main() {
int n; int q;
while(~scanf("%d%d", &n, &q)) {
vector<int> a(n);
for(int i = 0; i < n; ++ i)
scanf("%d", &a[i]);
vector<int> sum(n + 1);
rep(i, n) sum[i + 1] = sum[i] ^ a[i];
vector<int> sum2(n + 2);
rep(i, n + 1) sum2[i + 1] = sum2[i] + sum[i];
rep(ii, q) {
int x; int y; int k;
scanf("%d%d%d", &x, &y, &k), ++ y;
int cnt0 = sum2[y + 1] - sum2[x];
int cnt1 = (y + 1 - x) - cnt0;
if(cnt0 > cnt1) swap(cnt0, cnt1);
ll ans = k == 0 ? (ll)cnt0 * cnt1 :
(ll)((y + 1 - x) / 2) * ((y + 1 - x + 1) / 2);
printf("%lld\n", ans);
}
}
return 0;
}
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
public class C {
BufferedReader br;
PrintWriter out;
StringTokenizer st;
boolean eof;
void solve() throws IOException {
int n = nextInt();
int q = nextInt();
int[] b = new int[n + 1];
for (int i = 0; i < n; i++) {
int x = nextInt();
b[i + 1] = b[i] ^ x;
}
int[] c = new int[n + 2];
for (int i = 0; i < n + 1; i++) {
c[i + 1] = c[i] + b[i];
}
while (q-- > 0) {
int x = nextInt();
int y = nextInt();
int k = nextInt();
int len = y - x + 2;
int ones;
if (k == 0) {
ones = c[y + 2] - c[x];
} else {
ones = len / 2;
}
out.println((long)ones * (len - ones));
}
}
C() 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 C();
}
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());
}
}
Solution in Python :
In Python3 :
#!/usr/bin/python3
def init(A,n):
retA = [0]*(n+1);
B = [0]*(n+1)
tmp = 0
for i in range(n):
tmp ^= A[i]
B[i+1] = tmp
retA[i+1] += retA[i] + tmp
return retA,B
if __name__ == "__main__":
n,q = list(map(int,input().rstrip().split()))
A = list(map(int,input().rstrip().split()))
newA, B = init( A, n )
while q > 0 :
x,y,k = list(map(int,input().rstrip().split()))
num = newA[y+1] - newA[x]
if B[x]:
num = (y + 1 - x) - num
size = y - x + 2
num = abs(int(size/2)) if k else num
print( num*(size-num) )
q-=1
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