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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.     int total = 0;
  70.     for(int i = 0 ; i<n ; i++){
  71.         cin>>A[i];
  72.         total += A[i];
  73.     }
  74.     int p4[n],p2[n];
  75.     p4[n-1] = A[n-1];
  76.     int sum = A[n-1];
  77.     for(int i = n-2 ; i>=0 ; i--){
  78.         sum += A[i];
  79.         p4[i] = min((p4[i+1]),sum);
  80.     }
  81.     sum = 0;
  82.     for(int i = 0 ; i<n ; i++){
  83.         p2[i] = min(A[i],(sum+A[i]));
  84.     }
  85.     int ans = LLONG_MAX;
  86.     ans = min(p2[n-1],p4[0]);
  87.     for(int i = 0 ; i<n-1 ; i++){
  88.         int d = min(0LL,p2[i])+min(0LL,p4[i+1]);
  89.         ans = min(ans,d);
  90.     }
  91.     cout<<total-(2*ans)<<endl;
  92. }
  93. signed main() {
  94.     ios::sync_with_stdio(false); cin.tie(NULL);
  95.     int t;
  96.     cin >> t;
  97.     while (t--) {
  98.         solve();
  99.     }
  100.     return 0;
  101. }
  102.  
Success #stdin #stdout 0.01s 5288KB
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https://ideone.com/Z0xkMd
language:
C++14 (gcc 8.3)
created:
4 days ago
visibility:
public

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