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  1. #include <iostream>
  2. #include <vector>
  3. #include <algorithm>
  4. using namespace std;
  5. using ll = long long;
  6.  
  7. int main(){
  8.     ios::sync_with_stdio(false);
  9.     cin.tie(nullptr);
  10.  
  11.     int t;
  12.     cin >> t;
  13.     while(t--){
  14.         ll n;
  15.         cin >> n;
  16.         ll suma = 1, sumb = 1;
  17.         // 將 vector 大小調整為 n+1，第 n 個作為哨兵預設值 n
  18.         vector<ll> multia(n+1, n), multib(n+1, n), numa(n), numb(n);
  19.         vector<char> doa(n), dob(n);
  20.  
  21.         for (int i = 0; i < n; i++){
  22.             cin >> doa[i] >> numa[i] >> dob[i] >> numb[i];
  23.         }
  24.  
  25.         // 從 n-1 到 0 更新 multia 與 multib
  26.         for (int i = n - 1; i >= 0; i--){
  27.             if(doa[i] == 'x' && dob[i] == 'x'){
  28.                 if(numa[i] > numb[i]){
  29.                     multia[i] = i;
  30.                     multib[i] = multib[i+1];
  31.                 }
  32.                 else if(numb[i] > numa[i]){
  33.                     multib[i] = i;
  34.                     multia[i] = multia[i+1];
  35.                 }
  36.                 else{
  37.                     multia[i] = multia[i+1];
  38.                     multib[i] = multib[i+1];
  39.                 }
  40.             }
  41.             else{
  42.                 if(doa[i] == 'x')
  43.                     multia[i] = i;
  44.                 else
  45.                     multia[i] = multia[i+1];
  46.  
  47.                 if(dob[i] == 'x')
  48.                     multib[i] = i;
  49.                 else
  50.                     multib[i] = multib[i+1];
  51.             }
  52.         }
  53.  
  54.         for (int i = 0; i < n; i++){
  55.             ll x = 0;
  56.             if(doa[i] == '+')
  57.                 x += numa[i];
  58.             else
  59.                 x += (numa[i] - 1) * suma;
  60.  
  61.             if(dob[i] == '+')
  62.                 x += numb[i];
  63.             else
  64.                 x += (numb[i] - 1) * sumb;
  65.  
  66.             if(i < n - 1){
  67.                 if(multia[i+1] < multib[i+1])
  68.                     suma += x;
  69.                 else
  70.                     sumb += x;
  71.             }
  72.             else{
  73.                 cout << suma + sumb + x << "\n";
  74.             }
  75.         }
  76.     }
  77.     return 0;
  78. }
Success #stdin #stdout 0.01s 5284KB
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stdin
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4
3
+ 4 x 2
x 3 x 3
+ 7 + 4
4
+ 9 x 2
x 2 x 3
+ 9 + 10
x 2 + 1
4
x 2 + 1
+ 9 + 10
x 2 x 3
+ 9 x 2
5
x 3 x 3
x 2 x 2
+ 21 + 2
x 2 x 3
+ 41 x 3
stdout
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https://ideone.com/KLAr22
language:
C++ (gcc 8.3)
created:
1 day ago
visibility:
public

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