結果
問題 | No.1930 XOR of Two Range |
ユーザー |
|
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 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 1 |
other | WA * 3 |
ソースコード
#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_boundconst 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 => 0void 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; // 1else if (r == 3) cout << 1 << nl;else if (r == 4) cout << 1 << nl; // 3 -> 1else 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();}}