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  1. #include<bits/stdc++.h>
  2. using namespace std;
  3. #define endl '\n'
  4. #define int long long int
  5. const int MOD = 1000000007;
  6. const int MOD2 = 998244353;
  7. const int INF = LLONG_MAX / 2;
  8. const int MAXN = 100000;
  9. int primes[1000000];
  10.  
  11. /*void seive() {
  12.     fill(primes, primes + 1000000, 1);
  13.     primes[0] = primes[1] = 0;
  14.     for (int i = 2; i * i < 1000000; i++) {
  15.         if (primes[i]) {
  16.             for (int j = i * i; j < 1000000; j += i) {
  17.                 primes[j] = 0;
  18.             }
  19.         }
  20.     }
  21. }
  22.  
  23. bool isPrime(int n) {
  24.     if (n <= 1) return false;
  25.     for (int i = 2; i * i <= n; i++) {
  26.         if (n % i == 0) return false;
  27.     }
  28.     return true;
  29. }
  30.  
  31. int gcd(int a, int b) {
  32.     if (a == 0) return b;
  33.     return gcd(b % a, a);
  34. }
  35.  
  36. int power(int a, int b, int mod) {
  37.     int res = 1;
  38.     a %= mod;
  39.     while (b > 0) {
  40.         if (b & 1) res = res * a % mod;
  41.         a = a * a % mod;
  42.         b >>= 1;
  43.     }
  44.     return res;
  45. }
  46.  
  47. // nCr % MOD for n < MOD
  48. int nCrModP(int n, int r) {
  49.     if (r > n) return 0;
  50.     if (r == 0 || r == n) return 1;
  51.  
  52.     int numerator = 1, denominator = 1;
  53.     for (int i = 0; i < r; i++) {
  54.         numerator = (numerator * (n - i)) % MOD;
  55.         denominator = (denominator * (i + 1)) % MOD;
  56.     }
  57.     return (numerator * power(denominator, MOD - 2, MOD)) % MOD;
  58. }
  59.  
  60. // Lucas's Theorem
  61. int lucas(int n, int r) {
  62.     if (r == 0) return 1;
  63.     return (lucas(n / MOD, r / MOD) * nCrModP(n % MOD, r % MOD)) % MOD;
  64. }*/
  65. void solve() {
  66.     int n;
  67.     cin>>n;
  68.     int A[n];
  69.     for(int i = 0 ; i<n ; i++){
  70.         cin>>A[i];
  71.     }
  72.     int ans1 = LLONG_MIN,ans2=LLONG_MIN,ans3=LLONG_MIN,ans4=LLONG_MIN;
  73.     int maxi = LLONG_MIN,mini1 = LLONG_MAX,mini2=LLONG_MAX;
  74.     int p = -1;
  75.     for(int i = 0 ; i<n ; i++){
  76.        maxi = max(maxi,(A[i]+i));
  77.     }
  78.     for(int i = 0 ; i<n ; i++){
  79.         int d = A[i]+i;
  80.         if(d<mini1){
  81.             p = i;
  82.             mini1 = d;
  83.         }
  84.     }
  85.     ans1 = max(ans1,(maxi-(2*mini1)));
  86.     maxi = LLONG_MIN,mini1 = LLONG_MIN,mini2=LLONG_MIN;
  87.     for(int i = 0 ; i<n ; i++){
  88.         maxi = max(maxi,(i-A[i]));
  89.     }
  90.     for(int i = 0 ; i<n ; i++){
  91.         int d = A[i]-i;
  92.         if(d>mini1){
  93.             mini1 = d;
  94.         }
  95.     }
  96.     ans2 = max(ans2,(maxi+(2*mini1)));
  97.     p = -1;
  98.     maxi = LLONG_MIN,mini1 = LLONG_MIN,mini2=LLONG_MIN;
  99.     for(int i = 0 ; i<n ; i++){
  100.         maxi = max(maxi,(A[i]-i));
  101.     }
  102.     for(int i = 0 ; i<n ; i++){
  103.         int d = i-A[i];
  104.         if(d>mini1){
  105.             mini1 = d;
  106.         }
  107.     }
  108.     ans3 = max(ans3,(maxi+(2*mini1)));
  109.     mini1 = LLONG_MIN,mini2 = LLONG_MIN;
  110.     maxi = LLONG_MAX;
  111.     for(int i = 0 ; i<n ; i++){
  112.         maxi = min(maxi,(A[i]+i));
  113.     }
  114.     for(int i = 0 ; i<n ; i++){
  115.         int d = A[i]+i;
  116.         if(d>mini1){
  117.             mini1 = d;
  118.         }
  119.     }
  120.     ans4 = max(ans4,((2*mini1)-maxi));
  121.     cout<<ans1<<" "<<ans2<<" "<<ans3<<" "<<ans4<<endl;
  122.     cout<<max({ans1,ans2,ans3,ans4})<<endl;
  123. }
  124. signed main() {
  125.     ios::sync_with_stdio(false); cin.tie(NULL);
  126.     //int t;
  127.     //cin >> t;
  128.     //while (t--) {
  129.         solve();
  130.     //}
  131.     return 0;
  132. }
  133.  
Success #stdin #stdout 0s 5328KB
comments (?)
stdin 
5
1 2 3 4 5
stdout 
7 1 -1 17
17
https://ideone.com/v4pR3x
language:
C++14 (gcc 8.3)
created:
5 days ago
visibility:
public

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