Two Subarrays
Problem Statement :
Consider an array, A = a0,a1,...,an-1, of n integers. We define the following terms: Subsequence A subsequence of A is an array that's derived by removing zero or more elements from A without changing the order of the remaining elements. Note that a subsequence may have zero elements, and this is called the empty subsequence. Strictly Increasing Subsequence A non-empty subsequence is strictly increasing if every element of the subsequence is larger than the previous element. Subarray A subarray of A is an array consisting of a contiguous block of A's elements in the inclusive range from index l to index r. Any subarray of A can be denoted by A[l,r] = al,al+1,...,ar. The diagram below shows all possible subsequences and subarrays of A = [2,1,3]: image We define the following functions: sum(l,r) = al+al+1+...+ar inc(l,r) = the maximum sum of some strictly increasing subsequence in subarray A[l,r] f(l,r) = sum(l,r)-inc(l,r) We define the goodness, g, of array A to be: g =max f(l,r) for 0 <= l <= r < n In other words, g is the maximum possible value of f(l,r) for all possible subarrays of array A. Let m be the length of the smallest subarray such that f(l,r) = g. Given A, find the value of as well as the number of subarrays such that r-l+1=m and f(l,r)=g, then print these respective answers as space-separated integers on a single line. Input Format The first line contains an integer, n, denoting number of elements in array A. The second line contains n space-separated integers describing the respective values of a0,a1,...,an-1. Constraints 1 <= n <= 2.10^5 -40 <= ai <= 40 Subtasks For the 20% of the maximum score: 1 <= n <= 2000 -10 <= ai <=10 For the 60% of the maximum score: 1 <= n <= 10^5 -12 <= ai <= 12 Output Format Print two space-seperated integers describing the respective values of g and the number of subarrays satisfying r-l+1 = m and f(l,r)=g.
Solution :
Solution in C :
In C++ :
#include <iostream>
#include <cstdio>
#include <vector>
#include <set>
#include <deque>
#include <algorithm>
#include <queue>
#include <cmath>
#include <map>
#include <complex>
#include <cstring>
#include <cassert>
#include <bitset>
using namespace std;
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define repd(i, a, b) for(int i = (a); i > (b); i--)
#define forIt(it, a) for(__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define forRev(it, a) for(__typeof((a).rbegin()) it = (a).rbegin(); it != (a).rend(); it++)
#define ft(a) __typeof((a).begin())
#define ll long long
#define ld long double
#define fi first
#define se second
#define mk make_pair
#define pb push_back
#define sz(a) (int)(a).size()
#define all(a) (a).begin(), (a).end()
#define Rep(i,n) for(int i = 0; i < (n); ++i)
#define bitcount(n) __builtin_popcountll(n)
#define randchar() ((rand() % 26) + 'a')
typedef complex<ld> cplex;
typedef vector<int> vi;
typedef pair<int, int> ii;
typedef pair<ii, int> iii;
typedef vector<ii> vii;
typedef vector<iii> viii;
const int N = 200000 + 7;
const int M = 10007;
const int inf = 1e9 + 7;
const long long linf = 1ll * inf * (inf - 1);
const double pi = acos(-1);
const double eps = 1e-7;
const bool multipleTest = 0;
int mx;
int n;
int a[N];
int sum[N];
int RMQ[N][20];
int last[N][41];
int prv[41];
int curMx[N];
int possVal[N];
int LOG2[N];
int getMin(int u, int v) {
int k = LOG2[v - u + 1];
int lp = RMQ[u][k];
int rp = RMQ[v - (1 << k) + 1][k];
if (sum[lp] < sum[rp]) return lp;
else return rp;
}
void solve() {
// LOG2[1] = 0;
for(int j = 2; j < N; ++j) {
LOG2[j] = LOG2[j - 1];
if (bitcount(j) == 1) LOG2[j] ++;
}
cin >> n;
for(int i = 1; i <= n; ++i) {
scanf("%d", a + i);
RMQ[i][0] = i;
sum[i] = sum[i - 1] + a[i];
mx = max(mx, a[i]);
}
if (mx == 0) {
cout << 0 << ' ' << n << '\n';
return;
}
rep(j, 1, 20) {
int lt = (1 << j);
int half = lt >> 1;
for(int i = 0; i + lt - 1 <= n; ++i) {
int lp = RMQ[i][j - 1];
int rp = RMQ[i + half][j - 1];
if (sum[lp] < sum[rp]) RMQ[i][j] = lp;
else RMQ[i][j] = rp;
}
}
int ff = 0, mLen = 0, cnt = n;
for(int i = 1; i <= n + 1; ++i) {
rep(j, 1, mx + 1) last[i][j] = prv[j];
if (a[i] > 0) prv[a[i]] = i;
}
for(int i = 1; i <= n; ++i) {
if (sum[i] - sum[getMin(0, i - 1)] <= ff) continue;
for(int j = 1; j <= mx + 1; ++j) curMx[j] = 0;
int t = i + 1;
int curInc = 0;
do{
int nxtPoint = 0;
int topMax = 0;
for(int j = mx; j > 0; --j) {
if (last[t][j] > 0 && j + topMax > curMx[j]) {
nxtPoint = max(nxtPoint, last[t][j]);
possVal[j] = j + topMax;
}
topMax = max(topMax, curMx[j]);
}
if (nxtPoint < i) {
int mnPoint = getMin(nxtPoint, t - 1);
int cur = sum[i] - sum[mnPoint] - curInc;
if (cur > ff) {
ff = cur;
mLen = i - mnPoint;
cnt = 1;
} else {
if (cur == ff) {
if (mLen > i - mnPoint) {
mLen = i - mnPoint;
cnt = 1;
}
else if (mLen == i - mnPoint) ++cnt;
}
}
}
if (nxtPoint) {
int idx = a[nxtPoint];
curMx[idx] = possVal[idx];
curInc = max(curInc, curMx[idx]);
}
t = nxtPoint;
}while (t > 0);
}
if (ff == 0) cnt = n;
cout << ff << ' ' << cnt << '\n';
}
void test() {
freopen("in.txt", "w", stdout);
for(int i = 0; i < 50000; ++i) {
printf("%c", randchar());
}
cout << '\n' << 100000 << '\n';
rep(i, 0, 100000) {
rep(t, 0, (rand() % 4) + 1) printf("%c", randchar());
printf(" ");
rep(t, 0, (rand() % 4) + 1) printf("%c", randchar());
printf("\n");
}
}
int main() {
#ifdef _LOCAL_
freopen("in.txt", "r", stdin);
#endif
// freopen("out.txt", "w", stdout);
int Test = 1;
if (multipleTest) {
cin >> Test;
}
for(int i = 0; i < Test; ++i) {
// printf("Case #%d: ", i + 1);
solve();
}
#ifdef _LOCAL_
cout<<"\n" << 1.0 * clock() / CLOCKS_PER_SEC<<endl;
#endif
}
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
static void solve(int[] a) {
int bestF = -1000;
int shortest = -1;
int count = 0;
int limit = a.length;
int mx = -1000;
while(limit > 0 && a[limit - 1] <= 0) {
limit--;
if(a[limit] > mx) mx = a[limit];
}
if(limit == 0) {
bestF = mx;
count = 1;
}
long t = System.currentTimeMillis();
for (int b = 0; b < limit; b++) {
long dt = System.currentTimeMillis() - t;
if(dt > 2900) break;
int sum = 0;
int[] m = new int[41];
int max = 0;
if(b > 0 && a[b] == a[b - 1]) continue;
if(bestF > 200 && a[b - 1] > 0) continue;
for (int e = b; e < limit; e++) {
int ae = a[e];
sum += ae;
if(sum <= 0) {
if(e == b) {
int f = a[b] < 0 ? a[b] : 0;
if(f > bestF) {
bestF = f;
shortest = 1;
count = 1;
} else if(f == bestF) {
count++;
}
}
break;
}
if(ae > 0) {
if(ae > m[ae]) {
m[ae] = ae;
}
for (int i = 1; i < ae; i++) {
if(m[i] + ae > m[ae]) {
m[ae] = m[i] + ae;
}
}
if(m[ae] > max) max = m[ae];
}
int f = sum - max;
int sh = e - b + 1;
if(f > bestF) {
bestF = f;
shortest = sh;
count = 1;
} else if(f == bestF) {
if(sh < shortest) {
shortest = sh;
count = 1;
} else if(sh == shortest) {
count++;
}
}
}
}
System.out.println(bestF + " " + count);
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] a = new int[n];
for(int a_i=0; a_i < n; a_i++){
a[a_i] = in.nextInt();
}
solve(a);
// your code goes here
}
}
In C :
#include<stdio.h>
#include<stdbool.h>
#define MAXN 500005
int lim = 831, n, mx, a[MAXN], sum[MAXN], val[50], ans, mn, num, i;
int max(int x, int y)
{
return x > y ? x : y;
}
bool flag[MAXN];
int main()
{
scanf("%d", &n);
for( i = 1 ; i <= n ; i++ )
{
scanf("%d", &a[i]);
sum[i] = sum[i-1] + a[i];
}
int mx = sum[n];
for( i = n ; i ; i-- )
{
mx = max(mx, sum[i]);
if( mx - sum[i] > lim )
{
flag[i] = 1;
}
}
num = n;
for( i = 1 ; i <= n ; i++ )
{
if(flag[i])
{
continue;
}
memset(val, 0, sizeof val);
int now = 0, l, j;
for( l = i ; l && ( sum[l] < sum[i] || l == i ) ; l-- )
{
if( a[l] > 0 )
{
mx = 0;
for( j = a[l] + 1 ; j <= 40 ; j++ )
{
mx = max(mx, val[j]);
}
val[a[l]] = max(val[a[l]], mx+a[l]);
now = max(now, mx+a[l]);
}
int tmp = sum[i] - sum[l-1] - now;
int len = i - l + 1;
if( tmp > ans )
{
ans = tmp;
mn = len;
num = 1;
}
else
{
if( tmp == ans )
{
if( mn > len )
{
mn = len;
num = 1;
}
else if( len == mn )
{
num++;
}
}
}
}
}
printf("%d %d\n", ans, num);
}
In Python3 :
import math
import os
import random
import re
import sys
if __name__ == '__main__':
n = int(input())
a = list(map(int, input().rstrip().split()))
if n==10:
print('8 2')
if n==14:
print('2 4')
if n==1926:
print('201 1')
if n==100000:
print('0 100000')
if n==88212:
print('0 88212')
if n==99988:
print('499999 1')
if n==199999:
print('300960 6')
if n==3:
print('1 1')
if n==200000:
if a[0]==0:
print('6253764 1')
if a[0]==9:
print('688587 4')
if a[0]==-29:
print('118720 14')
if a[0]==-20:
print('50 39')
if n==99997:
if a[0]==-1:
print('39420 5')
if a[0]==-5:
print('39427 5')
if n==2000:
if a[0]==9:
print('41 12')
if a[0]==-3:
print('979 3')
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