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#227637#7613. Inverse Problemucup-team1004ML 4ms57384kbC++144.3kb2023-10-27 20:19:502023-10-27 20:19:51

Judging History

你现在查看的是最新测评结果

  • [2023-10-27 20:19:51]
  • 评测
  • 测评结果:ML
  • 用时:4ms
  • 内存:57384kb
  • [2023-10-27 20:19:50]
  • 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;
}
ll f[N][N][N],g[N][N][N],F[N][N],G[N][N];
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;
	for(int lim=1;lim<=n;lim++){
		f[lim][0][0]=1;
		for(int r=1;r<=lim;r++){
			memcpy(f[lim][r],f[lim][r-1],sizeof f[lim][r]);
			for(int j=r;j<=lim;j++){
				f[lim][r][j]+=f[lim][r][j-r];
			}
			F[lim][r]=accumulate(f[lim][r],f[lim][r]+1+lim,0ll);
		}
	}
	for(int lim=1;lim<=n;lim++){
		g[lim][lim+1][0]=1;
		for(int r=lim;r>=1;r--){
			memcpy(g[lim][r],g[lim][r+1],sizeof g[lim][r]);
			for(int j=r;j<=lim;j++){
				g[lim][r][j]+=g[lim][r][j-r];
			}
			G[lim][r]=accumulate(g[lim][r],g[lim][r]+1+lim,0ll);
		}
	}
	// for(int lim=1;lim<=n;lim++){
	// 	debug("F",ary(F[lim],1,lim));
	// 	debug("G",ary(G[lim],1,lim));
	// }
}
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<pair<int,vector<int> > >S[N],T[N];
vector<int>now;
void dfs1(int n,int l,int r,int sum,int mul){
	// if(sum<n&&sum+l>n)return;
	S[sum].push_back({mul,now});
	for(int i=l;i<=r&&sum+i<=n;i++){
		now.push_back(i);
		dfs1(n,i,r,sum+i,1ll*mul*ival[i]%mod);
		now.pop_back();
	}
}
void dfs2(int n,int l,int r,int sum,int mul){
	T[sum].push_back({mul,now});
	for(int i=l;i<=r&&sum+i<=n;i++){
		now.push_back(i);
		dfs2(n,i,r,sum+i,1ll*mul*val[i]%mod);
		now.pop_back();
	}
}
bitset<mod>vis;
int t3;
void solve(int n){
	int B=0;
	ll mn=1e18;
	for(int i=1;i<n;i++){
		ll now=max(F[n][i]*cnt,G[n][i+1]);
		if(now<mn)mn=now,B=i;
	}
	for(int i=1;i<=n;i++)val[i]=P(n,i),ival[i]=iP(n,i);
	for(int i=0;i<=n;i++)T[i].clear(),S[i].clear();
	clock_t st=clock();
	dfs1(n,1,B,0,1);
	dfs2(n,B+1,n,0,1);
	int t1=0,t2=0;
	for(int i=0;i<=n;i++)t1+=S[i].size(),t2+=T[i].size();
	debug("n",1.0*(clock()-st)/CLOCKS_PER_SEC,n,t1,t2,F[n][B],G[n][B+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 l=0;l<=n;l++){
		int r=n-l;
		for(auto &Tx:T[r]){
			vis[Tx.first]=1;
			t2++;
		}
		for(int i=1;i<=m;i++)if(!ans[i].first){
			int x=1ll*goal[i]*inv[n+1]%mod*inv[n+2]%mod;
			for(auto &Sx:S[l]){
				int mul=1ll*x*Sx.first%mod;
				t1++;
				if(!vis[mul])continue;
				// debug("???");
				for(auto &Tx:T[r]){
					t3++;
					if(Tx.first!=mul)continue;
					ans[i].first=n+2;
					vector<int>res(Sx.second);
					for(int x:Tx.second)res.push_back(x);
					res[0]++;
					for(;res.size()<n+2;res.push_back(0));
					ans[i].second=res,cnt--;
					debug("ok",i,ans[i].first,ans[i].second);
					break;
				}
				break;
			}
		}
		for(auto &Tx:T[r]){
			vis[Tx.first]=0;
		}
	}
}
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}};
		if(ans[i].first){
			debug("ok",i,ans[i].first,ans[i].second);
		}
	}
	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);
			}
		}
	}
	debug(1.0*clock()/CLOCKS_PER_SEC);
	return 0;
}

详细

Test #1:

score: 100
Accepted
time: 4ms
memory: 57384kb

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
Memory Limit Exceeded

input:

9
185396120
468170792
837583517
696626231
338497514
762842660
800028852
928391161
733524004

output:

14
1 2
1 3
2 4
3 5
4 6
5 7
6 8
7 9
8 10
9 11
9 12
10 13
10 14
122
1 2
1 3
2 4
3 5
4 6
5 7
6 8
6 9
7 10
7 11
8 12
8 13
9 14
9 15
9 16
10 17
10 18
10 19
11 20
11 21
11 22
12 23
12 24
12 25
12 26
13 27
13 28
13 29
13 30
14 31
14 32
14 33
14 34
14 35
15 36
15 37
15 38
15 39
15 40
15 41
15 42
16 43
16 44...

result: