QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#138490#6512. Completely Multiplicative Functionnikolapesic2802WA 78ms46812kbC++144.6kb2023-08-11 20:20:042023-08-11 20:20:06

Judging History

你现在查看的是最新测评结果

  • [2023-08-11 20:20:06]
  • 评测
  • 测评结果:WA
  • 用时:78ms
  • 内存:46812kb
  • [2023-08-11 20:20:04]
  • 提交

answer

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>

#define ll long long
#define pb push_back
#define f first
#define s second
#define sz(x) (int)(x).size()
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define ios ios_base::sync_with_stdio(false);cin.tie(NULL)
#define ld long double
#define li __int128

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>; ///find_by_order(),order_of_key()
template<int D, typename T>struct vec : public vector<vec<D - 1, T>> {template<typename... Args>vec(int n = 0, Args... args) : vector<vec<D - 1, T>>(n, vec<D - 1, T>(args...)) {}};
template<typename T>struct vec<1, T> : public vector<T> {vec(int n = 0, T val = T()) : vector<T>(n, val) {}};
template<class T1,class T2> ostream& operator<<(ostream& os, const pair<T1,T2>& a) { os << '{' << a.f << ", " << a.s << '}'; return os; }
template<class T> ostream& operator<<(ostream& os, const vector<T>& a){os << '{';for(int i=0;i<sz(a);i++){if(i>0&&i<sz(a))os << ", ";os << a[i];}os<<'}';return os;}
template<class T> ostream& operator<<(ostream& os, const deque<T>& a){os << '{';for(int i=0;i<sz(a);i++){if(i>0&&i<sz(a))os << ", ";os << a[i];}os<<'}';return os;}
template<class T> ostream& operator<<(ostream& os, const set<T>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T> ostream& operator<<(ostream& os, const set<T,greater<T> >& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T> ostream& operator<<(ostream& os, const multiset<T>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T> ostream& operator<<(ostream& os, const multiset<T,greater<T> >& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T1,class T2> ostream& operator<<(ostream& os, const map<T1,T2>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
int ri(){int x;scanf("%i",&x);return x;}
void rd(int&x){scanf("%i",&x);}
void rd(long long&x){scanf("%lld",&x);}
void rd(double&x){scanf("%lf",&x);}
void rd(long double&x){scanf("%Lf",&x);}
void rd(string&x){cin>>x;}
void rd(char*x){scanf("%s",x);}
template<typename T1,typename T2>void rd(pair<T1,T2>&x){rd(x.first);rd(x.second);}
template<typename T>void rd(vector<T>&x){for(T&p:x)rd(p);}
template<typename C,typename...T>void rd(C&a,T&...args){rd(a);rd(args...);}
//istream& operator>>(istream& is,__int128& a){string s;is>>s;a=0;int i=0;bool neg=false;if(s[0]=='-')neg=true,i++;for(;i<s.size();i++)a=a*10+s[i]-'0';if(neg)a*=-1;return is;}
//ostream& operator<<(ostream& os,__int128 a){bool neg=false;if(a<0)neg=true,a*=-1;ll high=(a/(__int128)1e18);ll low=(a-(__int128)1e18*high);string res;if(neg)res+='-';if(high>0){res+=to_string(high);string temp=to_string(low);res+=string(18-temp.size(),'0');res+=temp;}else res+=to_string(low);os<<res;return os;}

const int N=1e6+5;
vector<int> di(N);
vector<bool> zavisi(N);
vector<int> od[N];
vector<int> val(N);
void test(){
	int n=ri(),k=ri();
	if((n&1)!=(k&1)){
		printf("-1\n");
		return;
	}
	if(n==1){
		printf("1\n");
		return;
	}
	for(int i=1;i<=n;i++){
		val[i]=1;
	}
	int sum=n;
	for(int i=2;i<=n;i++){
		//printf("%i: mogu %i\n",i,sum-2*imam);
		if(sum==k){
			for(int i=1;i<=n;i++){
				printf("%i ",val[i]);
			}
			printf("\n");
			return;
		}
		if(zavisi[i]!=0){
			continue;
		}
		assert(val[i]==1);
		int sm=-2;
		for(auto p:od[i]){
			if(p>n){
				break;
			}
			if(val[p]==1){
				sm-=2;
			}
			else{
				sm+=2;
			}
		}
		//printf("%i: %i\n",i,sm);
		if(sm<0&&sum+sm>=k){
			sum+=sm;
			val[i]=-1;
			for(auto p:od[i]){
				if(p>n){
					break;
				}
				val[p]*=-1;
			}
		}
	}
	printf("-1\n");
}
int main()
{
	di[1]=1;
	for(int i=2;i<N;i++){
		if(di[i]==0){
			for(int j=i;j<N;j+=i){
				di[j]=i;
			}
		}
	}
	for(int i=2;i<N;i++){
		if(di[i]!=i){
			int lst=N,cnt=0;
			int tr=i;
			while(tr!=1){
				if(lst!=di[tr]){
					if(cnt&1){
						od[lst].pb(i);
					}
					lst=di[tr];
					cnt=1;
				}
				else{
					cnt++;
				}
				tr/=di[tr];
			}
			if(cnt&1){
				od[lst].pb(i);
			}
			zavisi[i]=1;
		}
	}
	int t=ri();
	while(t--){
		test();
	}
	return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 44ms
memory: 46760kb

input:

4
4 2
10 0
10 1
10 10

output:

1 -1 1 1 
1 -1 1 1 1 -1 -1 -1 1 -1 
-1
1 1 1 1 1 1 1 1 1 1 

result:

ok ok (4 test cases)

Test #2:

score: -100
Wrong Answer
time: 78ms
memory: 46812kb

input:

11475
1 0
1 1
2 0
2 1
2 2
3 0
3 1
3 2
3 3
4 0
4 1
4 2
4 3
4 4
5 0
5 1
5 2
5 3
5 4
5 5
6 0
6 1
6 2
6 3
6 4
6 5
6 6
7 0
7 1
7 2
7 3
7 4
7 5
7 6
7 7
8 0
8 1
8 2
8 3
8 4
8 5
8 6
8 7
8 8
9 0
9 1
9 2
9 3
9 4
9 5
9 6
9 7
9 8
9 9
10 0
10 1
10 2
10 3
10 4
10 5
10 6
10 7
10 8
10 9
10 10
11 0
11 1
11 2
11 3
11...

output:

-1
1
-1
-1
1 1 
-1
1 -1 1 
-1
1 1 1 
1 -1 -1 1 
-1
1 -1 1 1 
-1
1 1 1 1 
-1
1 -1 -1 1 1 
-1
1 -1 1 1 1 
-1
1 1 1 1 1 
1 -1 1 1 -1 -1 
-1
1 -1 1 1 1 -1 
-1
1 1 1 1 -1 1 
-1
1 1 1 1 1 1 
-1
1 -1 1 1 -1 -1 1 
-1
1 -1 1 1 1 -1 1 
-1
1 1 1 1 -1 1 1 
-1
1 1 1 1 1 1 1 
1 -1 1 1 -1 -1 1 -1 
-1
1 -1 1 1 1 -1...

result:

wrong answer ans finds the answer, but out doesn't (test case 3)