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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 b;
  67.     cin>>b;
  68.     vector<int>dp1(b+1,0),dp2(b+1,0);
  69.     dp1[0] = 1;
  70.     dp2[0] = 1;
  71.     for(int i = 1 ; i<=b ; i++){
  72.         char ch1;
  73.         cin>>ch1;
  74.         if(ch1=='+'){
  75.             int r;
  76.             cin>>r;
  77.             dp1[i] = max({(dp1[i-1]+r),(dp2[i-1]+r),(dp1[i-1])});
  78.             dp2[i] = min({(dp1[i-1]+r),(dp2[i-1]+r),(dp2[i-1])});
  79.         }
  80.         else if(ch1=='*'){
  81.             int r;
  82.             cin>>r;
  83.             dp1[i] = max({(dp1[i-1]*r),(dp2[i-1]*r),(dp1[i-1])});
  84.             dp2[i] = min({(dp1[i-1]*r),(dp2[i-1]*r),(dp2[i-1])});
  85.         }
  86.         else if(ch1=='-'){
  87.             int r;
  88.             cin>>r;
  89.             dp1[i] = max({(dp1[i-1]-r),(dp2[i-1]-r),(dp1[i-1])});
  90.             dp2[i] = min({(dp1[i-1]-r),(dp2[i-1]-r),(dp2[i-1])});
  91.         }
  92.         else if(ch1=='/'){
  93.             int r;
  94.             cin>>r;
  95.             dp1[i] = max({(dp1[i-1]/r),(dp2[i-1]/r),(dp1[i-1])});
  96.             dp2[i] = min({(dp1[i-1]/r),(dp2[i-1]/r),(dp2[i-1])});
  97.         }
  98.         else if(ch1=='N'){
  99.             dp1[i] = max({(-1*dp1[i-1]),(-1*dp2[i-1]),(dp1[i-1])});
  100.             dp2[i] = min({(-1*dp1[i-1]),(-1*dp2[i-1]),(dp2[i-1])});
  101.         }
  102.     }
  103.     cout<<max(dp1[b],dp2[b])<<endl;
  104. }
  105.  
  106. signed main() {
  107.     ios::sync_with_stdio(false); cin.tie(NULL);
  108.     //int t;
  109.     //cin >> t;
  110.     //while (t--) {
  111.         solve();
  112.     //}
  113.     return 0;
  114. }
  115.  
Success #stdin #stdout 0s 5312KB
comments (?)
stdin 
2
3
N
- 2
N
3
- 1
* 4
/ 2
stdout 
0
https://ideone.com/OpVz0L
language:
C++14 (gcc 8.3)
created:
24 days ago
visibility:
public

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