Iterate It
Problem Statement :
Consider the following pseudocode, run on an array of length :
rep := 0
while A not empty:
B := []
for x in A, y in A:
if x != y: append absolute_value(x - y) to B
A := B
rep := rep + 1
Given the values of and array , compute and print the final value of after the pseudocode above terminates; if the loop will never terminate, print -1 instead.
Input Format
The first line contains a single integer, , denoting the length of array .
The second line contains space-separated integers describing the respective values of .
Output Format
Print the final value of after the pseudocode terminates; if the loop will never terminate, print -1 instead.
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <stdbool.h>
#include <assert.h>
typedef unsigned int uint;
#define MAX_N 100000
#define MAX_VALUE 50000
uint a[MAX_N];
bool b[MAX_VALUE+1];
int main() {
// read input
uint n;
scanf("%u", &n);
assert(n <= MAX_N);
for (int i = 0; i < n; i++) {
uint v;
scanf("%u", &v);
assert(v);
assert(v <= MAX_VALUE);
b[v] = true;
}
// start grinding
uint rep = 0;
while (true) {
// transfer from b (presence array) back to a (sorted list)
uint stride = 0;
bool in_stride = true;
n = 0;
for (uint i = 1; i <= MAX_VALUE; i++) {
if (b[i]) {
b[i] = false;
if (!n) {
stride = i;
} else if (in_stride && i - a[n-1] != stride) {
in_stride = false;
}
a[n++] = i;
}
}
if (!n) {
break;
}
if (in_stride) {
// shortcut
assert(a[n-1]/stride == n);
rep += n;
break;
}
rep++;
for (uint ai = 0; ai < n-1; ai++) {
for (uint aj = ai+1; aj < n; aj++) {
// no unnecessary code here, performance-critical
b[a[aj] - a[ai]] = true;
}
}
}
printf("%u\n", rep);
return 0;
}
Solution in C++ :
In C ++ :
#include <iostream>
#include <cstdio>
#include <string.h>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <sstream>
#include <cmath>
#include <ctime>
#include <bitset>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector< vector<int> > vvi;
typedef vector<ll> vl;
typedef vector< vector<ll> > vvl;
#define forn(i, n) for (int i = 0; i < (int)(n); i++)
#define forv(i, v) forn(i, v.size())
#define all(v) v.begin(), v.end()
#define mp make_pair
#define pb push_back
const int N = 50005;
typedef bitset<N> mask;
int gcd(int x, int y) {
if (!y) return x;
return gcd(y, x % y);
}
int main() {
#ifdef NEREVAR_PROJECT
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
int n; cin >> n;
vi a(n);
forn(i, n) {
scanf("%d", &a[i]);
}
sort(all(a));
a.erase(unique(all(a)), a.end());
n = (int)a.size();
mask diffs;
mask present;
for (int i = n - 1; i >= 0; i--) {
diffs |= present >> a[i];
present.set(a[i]);
}
vi s;
forn(i, N) {
if (diffs.test(i)) {
s.pb(i);
}
}
if (s.empty()) {
cout << 1 << endl;
return 0;
}
int g = 0;
forv(i, s) g = gcd(g, s[i]);
forv(i, s) s[i] /= g;
diffs.reset();
present.reset();
for (int i = (int)s.size() - 1; i >= 0; i--) {
diffs |= present >> (s[i] + 1);
present.set(s[i]);
}
int steps = 2, m = s.back() - s[0];
while (m && !diffs.test(0)) {
mask next;
int mNext = m;
forn(i, m) {
if (diffs.test(i)) {
if (mNext == m) {
mNext = m - i - 1;
}
next |= diffs >> (i + 1);
}
}
m = mNext;
steps++;
diffs = next;
}
cout << steps + m << endl;
return 0;
}
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
private static final int l = 60000;
private static int gcd(int a, int b){
if (a < b) return gcd(b, a);
if (b == 0) return a;
return gcd(b, a % b);
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
boolean[] list = new boolean[l+1];
Set<Integer> set = new HashSet<Integer>();
for (int i = 0; i < n; i++){
int a = in.nextInt();
set.add(a);
list[a] = true;
}
boolean[] nList = new boolean[l+1];
for (int e : set){
for (int i = 1; i + e < l; i++){
nList[i] |= list[i + e];
/*
if (i < 10){
if ((list[i + e / b] >> (e % b)) > 0 || (list[i + e / b + 1] << (b - (e % b))) > 0){
System.out.println(bits(list[i + e / b] >> (e % b)));
System.out.println(bits(list[i + e / b + 1] << (b - (e % b))));
}
else{
System.out.println(bits(list[i + e / b]) + " " + (e % b));
System.out.println(bits(list[i + e / b + 1]) + " " + (b - (e % b)));
}
System.out.println(e + " " + i);
}*/
}
}
list = nList;
int g = 0;
int min = -1;
int max = 0;
//for (int a : set)
//System.out.println(a);
//System.out.println("-----");
set.clear();
for (int i = 0; i < l+1; i++){
if (list[i]){
//System.out.println(a);
set.add(i);
if (min < 0) min = i;
max = i;
g = gcd(i, g);
}
}
//System.out.println("-----");
//System.out.println(min);
//System.out.println(max);
//System.out.println(g);
int o = 1;
if (set.size() == 0){
System.out.println(o);
return;
}
Set<Integer> nSet = new HashSet<Integer>();
for (int a : set)
nSet.add(a / g);
set = nSet;
min /= g;
max /= g;
list = new boolean[l+1];
for (int a : set)
list[a] = true;
while (min > 1){
nList = new boolean[l+1];
for (int a = min; a <= max; a++){
if (list[a]){
for (int k = 1; k + a < l; k++){
nList[k] |= list[k + a];
}
}
}
list = nList;
max -= min;
for (int a = 1; a <= max; a++){
if (list[a]){
min = a;
break;
}
}
o++;
}
System.out.println(o + max);
}
/*
private static String bits(int i){
String s = "";
for (int j = b-1; j >= 0; j--)
s += (i & (1 << j)) > 0 ? 1 : 0;
return s;
}*/
}//[]{}
Solution in Python :
In Python3 :
input()
A = set([int(m) for m in input().strip().split()])
def DeltaBasis2(AList,P=False):
if type(AList) == set:
AList = sorted(list(AList))
if len(AList) == 1:
return 1
Count = 0
while len(AList)>0:
if len(AList) == 1:
return Count + 1
LCM = True
for i1 in AList[1:]:
if i1 % AList[0] != 0:
LCM = False
break
if LCM:
AList = [int(m/AList[0]) for m in AList]
if (AList[0] == 1 and AList[1] == 2):
return Count + AList[-1]
Delta = set()
if len(AList) < 100:
MaxWidth = len(AList) - 1
else:
MaxWidth = int(len(AList)**0.75//1)
for W in range(1,MaxWidth+1):
for i1 in range(len(AList)-W):
Delta.add(abs(AList[i1+W]-AList[i1]))
Delta = sorted(list(Delta))
AList = sorted(list(set([m for m in Delta] + [AList[-1] - m for m in AList[:-1]])))
if P:
print(AList2)
Count += 1
return Count
print(DeltaBasis2(A))
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