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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#227559 | #7613. Inverse Problem | ucup-team1004 | TL | 1ms | 3612kb | C++14 | 3.2kb | 2023-10-27 18:31:08 | 2023-10-27 18:31:08 |
Judging History
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