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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#227559#7613. Inverse Problemucup-team1004TL 1ms3612kbC++143.2kb2023-10-27 18:31:082023-10-27 18:31:08

Judging History

你现在查看的是最新测评结果

  • [2023-10-27 18:31:08]
  • 评测
  • 测评结果:TL
  • 用时:1ms
  • 内存:3612kb
  • [2023-10-27 18:31:08]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
template<typename T>
ostream& operator << (ostream &out,const vector<T> &x){
	if(x.empty())return out<<"[]";
	out<<'['<<x[0];
	for(int i=1,len=x.size();i<len;i++)out<<','<<x[i];
	return out<<']';
}
template<typename T1,typename T2,typename T3>
ostream& operator << (ostream &out,const tuple<T1,T2,T3> &x){
	return out<<'('<<get<0>(x)<<','<<get<1>(x)<<','<<get<2>(x)<<')';
}
template<typename T>
vector<T> ary(const T *a,int l,int r){
	return vector<T>{a+l,a+1+r};
}
template<typename T>
void debug(T x){
	cerr<<x<<endl;
}
template<typename T,typename ... S>
void debug(T x,S...y){
	cerr<<x<<' ',debug(y...);
}
const int N=150,M=15,mod=1e9+7;
int n,m,cnt,goal[N];
pair<int,vector<int> >ans[N];
int fac[N],ifac[N],inv[N];
ll qpow(ll x,ll y=mod-2){
	ll ans=1;
	for(;y;(x*=x)%=mod,y>>=1)if(y&1)(ans*=x)%=mod;
	return ans;
}
void init(int n=N-1){
	for(int i=inv[1]=fac[0]=1;i<=n;i++)fac[i]=1ll*fac[i-1]*i%mod;
	for(int i=2;i<=n;i++)inv[i]=1ll*inv[mod%i]*(mod-mod/i)%mod;
	for(int i=ifac[0]=1;i<=n;i++)ifac[i]=1ll*ifac[i-1]*inv[i]%mod;
}
int P(int n,int m){
	if(0>m||m>n)return 0;
	return 1ll*fac[n]*ifac[n-m]%mod;
}
int iP(int n,int m){
	if(0>m||m>n)return 0;
	return 1ll*ifac[n]*fac[n-m]%mod;
}
int val[N],ival[N];
vector<tuple<int,int,vector<int> > >S,T;
vector<int>now;
void dfs1(int n,int l,int r,int sum,int mul){
	if(l>r){
		return S.push_back({sum,mul,now});
	}
	dfs1(n,l+1,r,sum,mul);
	if(sum+l<=n){
		now.push_back(l);
		dfs1(n,l,r,sum+l,1ll*mul*ival[l]%mod);
		now.pop_back();
	}
}
void dfs2(int n,int l,int r,int sum,int mul){
	if(l>r){
		return T.push_back({sum,mul,now});
	}
	dfs2(n,l+1,r,sum,mul);
	if(sum+l<=n){
		now.push_back(l);
		dfs2(n,l,r,sum+l,1ll*mul*val[l]%mod);
		now.pop_back();
	}
}
void solve(int n){
	int B=sqrt(n);
	for(int i=1;i<=n;i++)val[i]=P(n,i),ival[i]=iP(n,i);
	S.clear(),T.clear();
	dfs1(n,1,B,0,1);
	dfs2(n,B+1,n,0,1);
	// debug(ary(val,1,n),ary(ival,1,n));
	sort(S.begin(),S.end());
	sort(T.begin(),T.end());
	// debug(S),debug(T);
	for(int i=1;i<=m;i++)if(!ans[i].first){
		int x=1ll*goal[i]*inv[n+1]%mod*inv[n+2]%mod;
		// if(i==2)debug("goal",x);
		for(auto &St:S){
			int sum=n-get<0>(St),mul=1ll*x*get<1>(St)%mod;
			// if(sum!=2||mul!=6)continue;
			// debug("query",sum,mul);
			auto it=lower_bound(T.begin(),T.end(),make_tuple(sum,mul,vector<int>()));
			// debug(get<0>(*it),get<1>(*it),get<2>(*it));
			if(it==T.end()||get<0>(*it)!=sum||get<1>(*it)!=mul)continue;
			ans[i].first=n+2;
			vector<int>res(get<2>(St));
			for(int x:get<2>(*it))res.push_back(x);
			res[0]++;
			for(;res.size()<n+2;res.push_back(0));
			ans[i].second=res,cnt--;
			// debug("ok",i);
			break;
		}
	}
}
int main(){
	cin>>m,init();
	for(int i=1;i<=m;i++)cin>>goal[i];
	cnt=m;
	for(int i=1;i<=m;i++){
		if(goal[i]==1)cnt--,ans[i]={1,{0}};
		else if(goal[i]==2)cnt--,ans[i]={2,{1,0}};
	}
	for(n=3;cnt;n++)solve(n-2);
	for(int i=1;i<=m;i++){
		// debug(ans[i].first,ans[i].second);
		int n=ans[i].first;
		printf("%d\n",n);
		for(int j=0,x=1;j<n;j++){
			for(;ans[i].second[j]--;){
				printf("%d %d\n",j+1,x+++1);
			}
		}
	}
	return 0;
}

Details

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Test #1:

score: 100
Accepted
time: 1ms
memory: 3612kb

input:

4
2
360
1
509949433

output:

2
1 2
5
1 2
1 3
2 4
2 5
1
10
1 2
1 3
2 4
3 5
4 6
5 7
6 8
7 9
8 10

result:

ok OK (4 test cases)

Test #2:

score: -100
Time Limit Exceeded

input:

9
185396120
468170792
837583517
696626231
338497514
762842660
800028852
928391161
733524004

output:


result: