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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 |
---|---|---|---|---|---|---|---|---|---|
#254150 | #7688. Alea Iacta Est | ucup-team134# | WA | 9ms | 7600kb | C++17 | 4.4kb | 2023-11-18 02:12:49 | 2023-11-18 02:12:49 |
Judging History
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;
int dv[N];
ll t;
int best;
int x,y;
int n,m;
vector<pair<int,int>> fac;
void rec(int i,ll a){
if(i==sz(fac)){
ll b=t/a;
//printf("%lld %lld\n",a,b);
if((a%n==0)||(b%n==0))
if(a+b<best){
best=a+b;
x=a;
y=b;
}
return;
}
for(int j=0;j<=fac[i].s;j++){
rec(i+1,a);
a*=fac[i].f;
}
}
int main()
{
for(int i=2;i<N;i++){
if(dv[i])continue;
for(int j=i;j<N;j+=i)dv[j]=i;
}
int tt=ri();
while(tt--){
n=ri(),m=ri();
if(n>m)swap(n,m);
map<int,int> fact;
int X=n;
while(X!=1){
fact[dv[X]]++;
X/=dv[X];
}
X=m;
while(X!=1){
fact[dv[X]]++;
X/=dv[X];
}
fac.clear();
for(auto p:fact){
fac.pb(p);
}
t=(ll)n*m;
best=n+m;
x=n;
y=m;
rec(0,1);
//printf("%i %i!\n",x,y);
if(best==n+m){
printf("0\n0\n");
}
else{
if(x%n==0){
swap(x,y);
}
printf("%i",x);
for(int i=1;i<=x;i++){
printf(" %i",i);
}
printf("\n");
int skip=x;
printf("%i",y);
for(int i=0;i<y/n;i++){
int st=1+skip*i;
for(int j=0;j<n;j++){
printf(" %i",st+j);
}
}
printf("\n");
}
}
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 8ms
memory: 7600kb
input:
3 2 8 1 9 2 9
output:
4 1 2 3 4 4 1 2 5 6 3 1 2 3 3 1 4 7 3 1 2 3 6 1 2 4 5 7 8
result:
ok Correct. (3 test cases)
Test #2:
score: -100
Wrong Answer
time: 9ms
memory: 7564kb
input:
1 40013 40013
output:
0 0
result:
wrong answer Integer parameter [name=n1] equals to 0, violates the range [1, 120039] (test case 1)