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  1.     #include <bits/stdc++.h>
  2.     #include <chrono>
  3.     using namespace std;
  4.     using namespace chrono;
  5.     // "AJEET JAIN"----"JAI JINENDRA"  
  6.     /* "णमो अरिहंताणं",
  7.         "णमो सिद्धाणं",
  8.         "णमो आयरियाणं",
  9.         "णमो उवज्झायाणं",
  10.         "णमो लोए सव्वसाहूणं",
  11.         "",
  12.         "एसो पंच नमोक्कारो, सव्व पावप्पणासणो",
  13.         "मंगलाणं च सव्वेसिं, पडमं हवै मंगलं",   */
  14.  
  15.  
  16.     // Aliases to op
  17.     using ll = long long;
  18.     using ull = unsigned long long;
  19.     using ld = double;
  20.     using vll = vector<ll>;
  21.  
  22.  
  23.     // Constants
  24.     constexpr ll INF = 4e18;
  25.     constexpr ld EPS = 1e-9;
  26.     constexpr ll MOD = 1e9 + 7;
  27.  
  28.  
  29.  
  30.     // Macros
  31.     #define F first
  32.     #define S second
  33.     #define all(x) begin(x), end(x)
  34.     #define allr(x) rbegin(x), rend(x)
  35.     #define py cout<<"YES\n";
  36.     #define pn cout<<"NO\n";
  37.     #define forn(i,n) for(int i=0;i<n;i++)
  38.     #define for1(i,n) for(int i=1;i<=n;i++)
  39.  
  40.     // #define insert push_back
  41.     #define pb push_back
  42.     #define MP make_pair
  43.     #define endl '\n'
  44.  
  45.     /*
  46.       remove substring or subarray ---> try to think about sliding w
  47.     
  48.     */                  
  49.  
  50.      /*
  51.       
  52.      Golden Rule
  53.  
  54.      1) problem is easy
  55.      2) proofs is easy
  56.      3) implementation is easy
  57.      
  58.      /*
  59.          ROUGH --
  60.  
  61.          ai*(j - 1) + bi*(n - j) = min
  62.            ai and bi for every i is constant we can play with position
  63.              expand form = ai*j - ai + bi*n -bi*j
  64.                  rearrange -> (ai - bi)*j + bi*n - ai
  65.                    sum of all i from 1 <= i <= n -->> ((a1 - b1)j1 + b1*n - a1) + ((a2 - b2)j2 + b1*n - a2)......+ ((an - bn)jn + bn*n - an)
  66.                      => (a1 - b1)j1 + (a2 - b2)j2 ... + (an - bn)jn + (b1 + b2 + .... bn) - (a1 + a2 + .... an)
  67.                          now just choose ji according to (ai - bi)
  68.                  
  69.      */
  70.  
  71.     void AJNJ(){
  72.        int n;
  73.        cin >> n;
  74.        int x = n;
  75.        vector<ll> v;
  76.        ll sum_a = 0 , sum_b = 0;
  77.        while(x--){
  78.           ll a , b;
  79.           cin >> a >> b;
  80.           sum_a += a; 
  81.           sum_b += b;
  82.           v.push_back(a - b);
  83.        }
  84.        sort(allr(v));
  85.        ll ans = 0;
  86.        for(int i = 0 ; i < v.size() ; i++){
  87.            ans += v[i]*(i + 1);
  88.        }
  89.  
  90.        cout << ans + ((n*sum_b) - sum_a) << endl;
  91.  
  92.     }
  93.  
  94.  
  95.     int main(){
  96.         ios::sync_with_stdio(0);
  97.         cin.tie(0);
  98.         cout.tie(0);
  99.         int T = 1;
  100.         // cin>>T;
  101.         auto start1 = high_resolution_clock::now();
  102.         while(T--){
  103.             AJNJ();
  104.         }
  105.         auto stop1 = high_resolution_clock::now();
  106.         auto duration = duration_cast<microseconds>(stop1 - start1);
  107.         cerr << "Time: " << duration . count() / 1000 << " ms" << endl;
  108.  
  109.         return 0;
  110.     }
Success #stdin #stdout #stderr 0.01s 5320KB
comments (?)
stdin 
Standard input is empty
stdout 
-3439090394622728964
stderr 
Time: 1 ms
https://ideone.com/Bnmb7x
language:
C++14 (clang 8.0)
created:
12 hours ago
visibility:
public

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