結果

問題 No.1930 XOR of Two Range
ユーザー Lê Hoàng NgôLê Hoàng Ngô
提出日時 2024-06-29 12:00:18
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 35 ms / 2,000 ms
コード長 3,201 bytes
コンパイル時間 1,426 ms
コンパイル使用メモリ 167,644 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-06-29 12:00:21
合計ジャッジ時間 2,255 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge4
このコードへのチャレンジ
（要ログイン）

テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 1 ms
6,816 KB
testcase_01 AC 25 ms
6,944 KB
testcase_02 AC 35 ms
6,944 KB
testcase_03 AC 35 ms
6,940 KB
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 # 

#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; // 4 -> 2
        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 == 6) cout << 0 << nl; // 6-> 4 -> 2 
        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();
    }
}
0