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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#93831#5521. Excellent XOR ProblemAsad_BinWA 2ms3508kbC++173.3kb2023-04-03 04:33:492023-04-03 04:33:52

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-04-03 04:33:52]
  • 评测
  • 测评结果:WA
  • 用时:2ms
  • 内存:3508kb
  • [2023-04-03 04:33:49]
  • 提交

answer

// . . . Bismillahir Rahmanir Rahim . . .
 
#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
#ifndef ONLINE_JUDGE
#define dbg_out cout
#define debug(...) dbg_out << "DBG )> "; __f(#__VA_ARGS__, __VA_ARGS__);
template<typename T1, typename T2> ostream& operator<<(ostream& out, pair<T1, T2> pr) { out << "{ " << pr.first << ", " << pr.second << " }"; return out; }
template<typename T1> ostream& operator<<(ostream& out, vector<T1> vec) { out << "{ "; for (auto &x: vec) out << x << ", "; out << "}"; return out; }
template<typename T1, size_t size> ostream& operator<<(ostream& out, array<T1, size> arr) { out << "{ "; for (auto &x: arr) out << x << ", "; out << "}"; return out; }
template<typename T1, typename T2> ostream& operator<<(ostream& out, map<T1, T2> mp) { out << "{ ";for (auto &x: mp) out << x.first << ": " << x.second <<  ", "; out << "}"; return out; }
template <typename Arg1> void __f(const char* name, Arg1&& arg1) { while (isspace(name[0])) name++; (isalpha(name[0]) || name[0] == '_') ? dbg_out << name << ": " << arg1 << "\n" : dbg_out << arg1 << "\n"; dbg_out.flush();}
template <typename Arg1, typename... Args> void __f (const char* names, Arg1&& arg1, Args&&... args) { const char *comma = strchr(names + 1, ','); while (isspace(names[0])) names++; (isalpha(names[0]) || names[0] == '_') ? dbg_out.write(names, comma - names) << ": " << arg1 << " | " : dbg_out << arg1 << " | "; __f(comma + 1, args...);}
#else
#define debug(...)
#endif
 
ll gcd(ll a, ll b){ while (b){ a %= b; swap(a, b);} return a;}
ll lcm(ll a, ll b){ return (a/gcd(a, b)*b);}
ll ncr(ll a, ll b){ ll x = max(a-b, b), ans=1; for(ll K=a, L=1; K>=x+1; K--, L++){ ans = ans * K; ans /= L;} return ans;}
ll bigmod(ll a,ll b,ll mod){ if(b==0){ return 1;} ll tm=bigmod(a,b/2,mod); tm=(tm*tm)%mod; if(b%2==1) tm=(tm*a)%mod; return tm;}
ll egcd(ll a,ll b,ll &x,ll &y){ if(a==0){ x=0; y=1; return b;} ll x1,y1; ll d=egcd(b%a,a,x1,y1); x=y1-(b/a)*x1; y=x1; return d;}
ll modpow(ll a,ll p,ll mod) {ll ans=1;while(p){if(p%2)ans=(ans*a)%mod;a=(a*a)%mod;p/=2;} return ans;}
ll inverse_mod(ll n,ll mod) {return modpow(n,mod-2,mod);}


int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	int t; cin >> t;
	while(t--){
		int n; cin >> n;
		
		int ara[n];
		for(int K = 0; K < n; K++) cin >> ara[K];
		
		int x_or = 0;
		for(int K = 0; K < n; K++) x_or ^= ara[K];
		
		if(x_or){
			int val = 0;
			bool ok = 1;
			for(int K = 0; K < n; K++){
				val ^= ara[K]; x_or ^= ara[K];
				
				if(val != x_or){
					cout << "YES\n";
					cout << 2 << "\n";
					cout << 1 << ' '<< K+1 << "\n";
					cout << K+2 << ' ' << n << "\n";
					ok = 0;
					break;
				}
			}
			
			if(ok) cout << "NO\n";
		}
		else{
			int val = 0;
			bool ok = 1;
			for(int K = 0; K < n-2; K++){
				if(!ara[K]) continue;
				x_or ^= ara[K];
				
				val = 0;
				for(int L = K+1; L < n; L++){
					val ^= ara[L]; x_or ^= ara[L];
					
					if(val != x_or && val != ara[K] && x_or != ara[K]){
						cout << "YES\n";
						cout << 2 << "\n";
						cout << 1 << ' '<< K+1 << "\n";
						cout << K+2 << ' ' << L+1 << "\n";
						cout << L+2 << ' ' << n << "\n";
						ok = 0;
						break;
					}
				}
				
				if(ok) cout << "NO\n";
				ok = 0;
				
				if(ara[K]) break;
			}
			if(ok) cout << "NO\n";
		}
	}
	
	return 0;
}
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

详细

Test #1:

score: 0
Wrong Answer
time: 2ms
memory: 3508kb

input:

4
2
0 0
3
1 2 3
5
16 8 4 2 1
6
42 42 42 42 42 42

output:

NO
YES
2
1 1
2 2
3 3
YES
2
1 1
2 5
NO

result:

FAIL has to be a partition (test case 2)