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  1.  
  2.     #include <bits/stdc++.h>
  3.  
  4.     using namespace std;
  5.  
  6.     using ll = long long;
  7.     const int MOD = 1000000007; 
  8.     const int MOD2 =  998244353; 
  9.     const ll INF = 1e18;
  10.     const int MX = 1000001; //check the limits, dummy
  11.  
  12.  
  13.     ll modExp(ll base, ll power) {
  14.         if (power == 0) {
  15.             return 1;
  16.         } else {
  17.             ll cur = modExp(base, power / 2); cur = cur * cur; cur = cur % MOD;
  18.             if (power % 2 == 1) cur = cur * base;
  19.             cur = cur % MOD;
  20.             return cur;
  21.         }
  22.     }
  23.  
  24.     ll inv(ll base) {
  25.         return modExp(base, MOD-2);
  26.     }
  27.  
  28.  
  29.     ll mul(ll A, ll B) {
  30.         return (A*B)%MOD;
  31.     }
  32.  
  33.     ll add(ll A, ll B) {
  34.         return (A+B)%MOD;
  35.     }
  36.  
  37.     ll dvd(ll A, ll B) {
  38.         return mul(A, inv(B));
  39.     }
  40.  
  41.     ll sub(ll A, ll B) {
  42.         return (A-B+MOD)%MOD;
  43.     }
  44.  
  45.     ll* facs = new ll[MX];
  46.     ll* facInvs = new ll[MX];
  47.  
  48.     ll choose(ll a, ll b) {
  49.         if (b > a) return 0;
  50.         if (a < 0) return 0;
  51.         if (b < 0) return 0;
  52.         ll cur = facs[a];
  53.         cur = mul(cur, facInvs[b]);
  54.         cur = mul(cur, facInvs[a-b]);
  55.         return cur;
  56.     }
  57.  
  58.  
  59.     void initFacs() {
  60.         facs[0] = 1; 
  61.         facInvs[0] = 1;
  62.         for (int i = 1 ; i < MX ; i ++ ) {
  63.             facs[i] = (facs[i-1] * i) % MOD;
  64.             facInvs[i] = inv(facs[i]);
  65.         }
  66.     }
  67.     // [1,2,3,4,5,6,7,8,9,10] 
  68.     // 
  69.     // how do you get the w
  70.     ll solve(ll left ,ll right, ll n,ll k ) {
  71.         int l = (right - left + 1);
  72.         if (l  <= k) {
  73.             return 0; 
  74.         }
  75.         ll res = 0;
  76.         ll mid = (right - left)/2; 
  77.         if (l % 2) {
  78.             ll k = ((n - mid)/l);
  79.             res += (mid * (k + 1)) + (l * (k * (k + 1))/2);
  80.             res += solve(1,mid - 1,n,k ); 
  81.         } else {
  82.             res += solve(1,mid,n,k); 
  83.         }
  84.         return res;
  85.     }
  86.     int main()  {
  87.         ios_base::sync_with_stdio(0); cin.tie(0);  
  88.         int t; cin >> t; 
  89.         while ( t --) {
  90.             ll n,k; cin >> n >> k;
  91.             cout << solve(1,n ,n, k) << endl; 
  92.         }
  93.  
  94.  
  95.  
  96.  
  97.  
  98.         return 0;
  99.     }
  100.  
Success #stdin #stdout 0.01s 5316KB
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stdin
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stdout
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3
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35904706226
https://ideone.com/gDLOZv
language:
C++ (gcc 8.3)
created:
18 hours ago
visibility:
public

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