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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.     void initFacs() {
  59.         facs[0] = 1; 
  60.         facInvs[0] = 1;
  61.         for (int i = 1 ; i < MX ; i ++ ) {
  62.             facs[i] = (facs[i-1] * i) % MOD;
  63.             facInvs[i] = inv(facs[i]);
  64.         }
  65.     }
  66.  
  67.     ll solve(ll l, ll r , ll n ,ll k) {
  68.         if ((r - l + 1)  < k ) { 
  69.             return 0; 
  70.         }        
  71.         ll res = 0 ; 
  72.         ll m = (l + r) / 2;
  73.         if ((r - l + 1) % 2) {
  74.  
  75.             res += m;
  76.             res += solve(l,   m-1, n, k);
  77.             res += solve(m+1, r,   n, k);
  78.         } else {
  79.             // (r-l+1) is odd → you do NOT add m
  80.             res += solve(l, m,   n, k);
  81.             res += solve(m+1, r, n, k);
  82.             // (r-l+1) is even → you incorrectly add m here
  83.         }
  84.  
  85.         return res; 
  86.     }
  87.  
  88.     int main()  {
  89.         ios_base::sync_with_stdio(0); cin.tie(0);  
  90.         int t; cin >> t;
  91.         while (t --) {
  92.             ll n ,k ; cin >> n >> k; 
  93.             ll res = solve(1,n ,n,k);
  94.  
  95.             cout << res << endl;
  96.  
  97.         }
  98.         return 0;
  99.     }
  100.  
Success #stdin #stdout 0.11s 5320KB
comments (?)
stdin 
6
7 2
11 3
55 13
5801 6
8919 64
8765432 1
stdout 
12
18
196
1975581
958900
38416403456028
https://ideone.com/Vo1mRb
language:
C++ (gcc 8.3)
created:
19 hours ago
visibility:
public

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