結果

問題 No.1930 XOR of Two Range
ユーザー Lê Hoàng Ngô
提出日時 2024-06-29 11:56:14
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,143 bytes
コンパイル時間 1,668 ms
コンパイル使用メモリ 168,548 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-06-29 11:56:17
合計ジャッジ時間 2,590 ms
ジャッジサーバーID
(参考情報)
judge4 / judge5
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 1
other WA * 3
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define all(x) x.begin(),x.end()
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define rep_r(i,a,b) for(int i=a;i>=b;i--)
#define each(a,x) for (auto& x : a)
using pi = pair<int,int>;
using pl = pair<ll,ll>;
using vi = vector<int>;
using vl = vector<ll>;
#define sz(x) int(x.size())
#define so(x) sort(all(x))
#define so_r(x) sort(all(x),greater<int>())
#define lb lower_bound
#define ub upper_bound
const char nl = '\n';
int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};
int bit_cnt(int x){
return __builtin_popcount(x);
}
ll bex(ll a, ll b, ll mod = 1e9 + 7){ll res = 1LL; while(b){ if (b&1) res = res * a % mod; a = a * a % mod; b >>= 1;} return res;}
template<class t,class u> bool chmax(t&a,u b){if(a<b){a=b;return true;}else return false;}
template<class t,class u> bool chmin(t&a,u b){if(b<a){a=b;return true;}else return false;}
const int MOD = 998244353;
const int N = 5e5 + 5;
const ll INF = 1e16;
// https://yukicoder.me/problems/no/1930
// T test case
// each test consist L,R
// output: xor of all (i + j) that i + j <= L + R and L <= i <= j <= R
// case L == R => ans = 2 * L = (L << 1)
// case L + 1 == R => ans = (2L) ^ (2L + 1) = 1
// case L + 2 == R => ans = (2L) ^ (2L + 1) ^ (2L + 2) ^ (2L + 2) = 1
// case L + 3 == R => ans = (2L) ^ (2L+1) ^ (2L+2) ^ (2L+3)
// ^ (2L+2) ^ (2L+3) = 1
// case L + 4 == R => ans = (2L) ^ (2L+1) ^ (2l+2) ^ (2L+3) ^ (2L+4)
// ^ (2L+2) ^ (2L+3) ^ (2L+4)
// ^ (2L+4) = 1 ^ (2L + 4) = 2L + 5
// 5 -> 3 -> 1 = 0 ^ 1 ^ (2L+4)
// case L + 5 == R => ans = 6 -> 4 -> 2 = 0
// case L + 6 == R => ans = 7 -> 5 -> 3 -> 1 = 0
// case L + 7 == R => ans = 8 -> 6 -> 4 -> 2 = 0
// 10 -> 8 -> 6 -> 4 -> 2 =
// 9 -> 7 -> 5 -> 3 -> 1 = 2L + 8 (diff == 9)
// 11 -> 9 -> 7 -> 5 -> 3 -> 1 = 1 ^ 1 ^ 1 = 1 (diff == 11)
// 13 -> 11 -> 9 -> 7 -> 5 -> 3 -> 1 = 1 ^ 1 ^ 1
// diff = R - L + 1
// if diff % 8 == 1,3,5,7
// = 1 => L + R
// = 3 => 1
// = 5 => L + R + 1
// = 7 => 0
void solve(){
int T; cin >> T;
while(T--){
ll L,R; cin >> L >> R;
ll diff = R - L + 1;
ll r = diff % 8;
if (r == 1) cout << L + R << nl;
else if (r == 2) cout << 1 << nl; // 1
else if (r == 3) cout << 1 << nl;
else if (r == 4) cout << 1 << nl; // 3 -> 1
else if (r == 5) cout << L + R + 1 << nl; // in this case: 1 ^ (L + R) = L + R + 1 (because l + r % 2 == 0)
else if (r == 7) cout << 0 << nl;
// example L + 6 == R => 7 -> 5 -> 3 -> 1 => 1 ^ 1 = 0 (if we add more 8 to R then we still have (1 ^ 1) = 0 ^ old = old)
else if (r == 0) cout << 0 << nl;
}
}
ll calc(ll l, ll r){
ll ans = 0;
rep(i,l,r){
rep(j,i,r){
if (i+j<=l+r) {
ans ^= (i + j);
}
}
}
return ans;
}
void test() {
rep(i,4,3+11) {
cout << calc(3,i) << nl;
}
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
// cin >> t;
// test();
while (t--){
solve();
}
}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0