prompt string | response string |
|---|---|
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include<iostream>
using namespace std;
int main(){
int N,d,p[1000],count=0;
cin >> N >> d;
for(int i=0;i<N;i++){
cin >> p[i];
if(p[i] > d){
count = count + (p[i]-d);
}
}
if(count == 0){
cout << "kusoge\n";
}
else{
cout << count << "\n";
}
return 0;
} |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.Closeable;
import java.io.FileInputStream;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.HashMap;
... |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include<bits/stdc++.h>
#define r(i,n) for(int i=0;i<n;i++)
using namespace std;
int main(){
int n,m,ans=0,x;
cin>>n>>m;
r(i,n){
cin>>x;
if(x>m)ans+=x-m;
}
if(!ans)cout<<"kusoge"<<endl;
else cout<<ans<<endl;
} |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include<bits/stdc++.h>
using namespace std;
int main(){
int N,d,p[1000]={},f=0,ans=0;
cin>>N>>d;
for(int i=0;i<N;i++)cin>>p[i];
for(int i=0;i<N;i++)
if(p[i]>d){
if(f==0)f=1;
ans+=p[i]-d;
}
if(f==1)cout<<ans<<endl;
else cout<<"kusoge"<<endl;
return 0;
} |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #define _USE_MATH_DEFINES
#include <cstdio>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <complex>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include... |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <cassert>
typedef long long ll;
using namespace std;
#define debug(x) c... |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include<iostream>
#include<sstream>
#include<fstream>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<algorithm>
#include<functional>
#include<utility>
#include<bitset>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<cstdio>
using names... |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include <bits/stdc++.h>
using namespace std;
int main()
{
int N, d;
cin >> N >> d;
int p[1001];
for(int i = 0; i < N; i++) cin >> p[i];
sort(p, p + N);
reverse(p, p + N);
int ans = -d + p[0];
for(int i = 1; i < N; i++) {
if(ans < ans + p[i] - d) ans += p[i] - d;
else break;
}
if(ans <= 0)... |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include<bits/stdc++.h>
using namespace std;
using Int = long long;
//INSERT ABOVE HERE
signed main(){
int n,d;
cin>>n>>d;
vector<int> p(n);
for(int i=0;i<n;i++) cin>>p[i];
int ans=0;
for(int i=0;i<n;i++) ans+=max(0,p[i]-d);
if(ans) cout<<ans<<endl;
else cout<<"kusoge"<<endl;
return 0;
}
|
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include <bits/stdc++.h>
#define rep(i, a, n) for(int i = a; i < n; i++)
#define repb(i, a, b) for(int i = a; i >= b; i--)
#define all(a) a.begin(), a.end()
#define int long long
using namespace std;
signed main(){
int n, d, a;
cin >> n >> d;
int ans = 0;
rep(i, 0, n){
cin >> a;
if(a > ... |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define ll long long
#define pp pair<ll,ll>
#define ld long double
#define all(a) (a).begin(),(a).end()
#define mk make_pair
ll MOD=998244353;
int inf=1000001000;
ll INF=1e18+5;
ll mod=INF;
int main() {
int n,d;
cin >>... |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include <iostream>
int main()
{
int N, d;
std::cin >> N >> d;
int p, res = 0;
for (int i = 0; i < N; i++) {
std::cin >> p;
res += std::max(0, p - d);
}
if (res == 0) {
std::cout << "kusoge" << std::endl;
} else {
std::cout << res << std::endl;
}
... |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;++i)
int main(void){
int N,d;
cin>>N>>d;
vector<int> p(N);
rep(i,N)cin>>p[i];
int res=0;
rep(i,N){
res+=max(0,p[i]-d);
}
if(res==0)cout<<"kusoge"<<endl;
else cout<<res<<endl;
return 0;
} |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include <bits/stdc++.h>
using namespace std;
#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define mp(a,b) make_pair((a),(b))
#define pb(a) push_back((a))
#define all(x) (x).begin(),(x).end()
#define uniq(x) sort(all(x)),(x).erase(unique(all(x)),end(x))
#define fi first
#define se... |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for (int i = 0; i < n; i++)
int main() {
int n, d, ans = 0;
cin >> n >> d;
REP(i,n){
int p;
cin >> p;
ans += max(0, p-d);
}
if(ans > 0){
cout << ans << endl;
}
else {
cout << "kusoge" <<... |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include<bits/stdc++.h>
using namespace std;
int main(){
int n,d,p[1010];
cin>>n>>d;
for(int i=0;i<n;i++)cin>>p[i];
sort(p,p+n);
int ans=0;
for(int i=0;i<n;i++){
if(p[n-1-i]<=d)break;
ans+=p[n-1-i]-d;
}
if(ans)cout<<ans<<endl;
else cout<<"kusoge"<<endl;
return 0;
}
|
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#define LP(i,n) for(int i=0;i<n;i++)
#define Mset(x,v) memset(x,v,sizeof(x))
#define ... |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include <iostream>
using namespace std;
int main(void){
int n,m,a,sum=0;
cin>>n>>m;
for(int i=0;i<n;i++){
cin>>a;
if(a>m)sum+=a-m;
}
if(sum<1)cout<<"kusoge"<<endl;
else cout<<sum<<endl;
return 0;
} |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include <iostream>
#include <algorithm>
using namespace std;
int n, p;
signed main() {
cin >> n >> p;
int ans = 0;
for (int i = 0; i < n; i++) {
int k; cin >> k;
ans += max(0, k - p);
}
if (!ans)cout << "kusoge\n";
else cout << ans << endl;
}
|
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <numeric>
#include <functional>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <sstream>
#include <string>
#define repd(i,a,b) for (int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repd(i,0,n... |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | #include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define all(a) (a).begin(),(a).end()
#define pb push_back
#define INF (1e9+1)
//#define INF (1LL<<59)
int main(){
int n,d;
cin>>n>>d;
vector<int> v(n);
rep(i,n)cin>>v[i];
i... |
Problem: B: Yamanote-line Game-Yamanote-line Game-
Bean (?) Knowledge
The Yamanote line is convenient. The reason is that you can ride as many laps as time allows just by paying 130 yen. However, if you board with a ticket, you must be careful about the valid time of the ticket. It seems safe with an IC card. Taking ... | import java.util.*;
public class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int d = sc.nextInt();
long total = 0;
for (int i = 0; i < n; i++) {
int x = sc.nextInt();
if (x >= d) {
total += x - d;
}
}
if (total <... |
Problem: G: Donuts Orientation
story
Homu-chan's recent boom is making sweets. It seems that he makes a lot of sweets and shares them with his friends. Homu-chan seems to be trying to make donuts this time.
There are many important things in making donuts. Not only the dough and seasoning, but also the decorations t... | #include <bits/stdc++.h>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd... |
Problem: G: Donuts Orientation
story
Homu-chan's recent boom is making sweets. It seems that he makes a lot of sweets and shares them with his friends. Homu-chan seems to be trying to make donuts this time.
There are many important things in making donuts. Not only the dough and seasoning, but also the decorations t... | #include<bits/stdc++.h>
#define X first
#define Y second
#define pb emplace_back
#define FOR(i,a,b) for(int (i)=(a);i<(b);++(i))
#define EFOR(i,a,b) for(int (i)=(a);i<=(b);++(i))
#define rep(X,Y) for (int (X) = 0;(X) < (Y);++(X))
#define reps(X,S,Y) for (int (X) = S;(X) < (Y);++(X))
#define rrep(X,Y) for (int (X) = (Y)... |
Problem: G: Donuts Orientation
story
Homu-chan's recent boom is making sweets. It seems that he makes a lot of sweets and shares them with his friends. Homu-chan seems to be trying to make donuts this time.
There are many important things in making donuts. Not only the dough and seasoning, but also the decorations t... | #include "bits/stdc++.h"
#include<vector>
#include<iostream>
#include<queue>
#include<algorithm>
#include<map>
#include<set>
#include<iomanip>
#include<assert.h>
#include<unordered_map>
#include<unordered_set>
#include<string>
#include<stack>
#include<complex>
#pragma warning(disable:4996)
using namespace std;
using ld... |
Problem: G: Donuts Orientation
story
Homu-chan's recent boom is making sweets. It seems that he makes a lot of sweets and shares them with his friends. Homu-chan seems to be trying to make donuts this time.
There are many important things in making donuts. Not only the dough and seasoning, but also the decorations t... | #include<iomanip>
#include<limits>
#include<thread>
#include<utility>
#include<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<numeric>
#include<cassert>
#include<random>
#include<chrono>
#include<unordered_set>
#includ... |
Problem: G: Donuts Orientation
story
Homu-chan's recent boom is making sweets. It seems that he makes a lot of sweets and shares them with his friends. Homu-chan seems to be trying to make donuts this time.
There are many important things in making donuts. Not only the dough and seasoning, but also the decorations t... | #include<bits/stdc++.h>
using namespace std;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
Int MOD;
template<typename T>
struct Mint{
T v;
Mint():v(0){}
Mint(signed v):v(v){}
Min... |
Problem: G: Donuts Orientation
story
Homu-chan's recent boom is making sweets. It seems that he makes a lot of sweets and shares them with his friends. Homu-chan seems to be trying to make donuts this time.
There are many important things in making donuts. Not only the dough and seasoning, but also the decorations t... | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef int _loop_int;
#define REP(i,n) for(_loop_int i=0;i<(_loop_int)(n);++i)
#define FOR(i,a,b) for(_loop_int i=(_loop_int)(a);i<(_loop_int)(b);++i... |
Problem: G: Donuts Orientation
story
Homu-chan's recent boom is making sweets. It seems that he makes a lot of sweets and shares them with his friends. Homu-chan seems to be trying to make donuts this time.
There are many important things in making donuts. Not only the dough and seasoning, but also the decorations t... | #include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
usi... |
Problem: G: Donuts Orientation
story
Homu-chan's recent boom is making sweets. It seems that he makes a lot of sweets and shares them with his friends. Homu-chan seems to be trying to make donuts this time.
There are many important things in making donuts. Not only the dough and seasoning, but also the decorations t... | // vvvvvvvvvvvv TEMPLATE vvvvvvvvvvvv
#include <bits/stdc++.h>
using namespace std; using ll = long long; using P = pair<ll, ll>;
const ll linf = 1e18; const double eps = 1e-12, pi = acos(-1);
#define FOR(i,a,b) for (ll i=(a),__last_##i=(b);i<__last_##i;i++)
#define RFOR(i,a,b) for (ll i=(b)-1,__last_##i=(a);i>=__last_... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> l_l;
typedef pair<l_l, l_l> llll;
typedef pair<int, int> i_i;
template<class T>
inline bool chmax(T &a, T b) {
if(a < b) {
a = b;
return true;
}
return false;
}
template<class T>
inline bool chmin(T &a,... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
int N;
long long dp[2][2020];
vector<pair<pair<long long,long long>,pair<long long,long long> > >E;
int main()
{
cin>>N;
for(int i=0;i<=2000;i++)dp[0][i]=-1e18;
for(int i=0;i<N;i++)
{
long long R,A,W,T;
cin>>R>>A>>W>>T;
T--;
E.push... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
const int INF = 1e8;
int main(){
int n;
cin>>n;
vector<int> r(n), a(n), w(n), t(n);
for( int i=0; i<n; ++i ){
cin>>r[i]>>a[i]>>w[i]>>t[i];
}
for( int i=0; i<n; ++i ){
int idx =... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include <bits/stdc++.h>
using namespace std;
const int inf=1e9;
struct data {
int r,a,w,t;
};
bool asc(const data &l,const data &r) {
return l.t+l.r>r.t+r.r;
}
int n;
vector<data> v;
int dp[2005][4005][2]={0};//dp[i][j][k][l]:i個目まで歌って容量がjでkのboolは歌ってるかどうか
//これだとそこのスタートのそれがそうなので無理
void chmax(int &a,int b){
a=max... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include <algorithm>
#include <bitset>
#include <cassert>
#include <chrono>
#include <climits>
#include <cmath>
#include <complex>
#include <cstring>
#include <deque>
#include <functional>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <s... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#include <tuple>
#define mkp make_pair
#define mkt make_tuple
#define rep(i,n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;
const ll ... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include<bits/stdc++.h>
#define ll long long
#define rep(i, n) for(int i=0;i<(n);++i)
#define per(i, n) for(int i=(n)-1;i>=0;--i)
#define repa(i, n) for(int i=1;i<(n);++i)
#define foreach(i, n) for(auto &i:(n))
#define pii pair<int, int>
#define pll pair<long long, long long>
#define all(x) (x).begin(), (x).end()
#defi... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include<bits/stdc++.h>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define HUGE_NUM 99999999999999999
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;
#define SIZE 2005
enum Type{
NOT,
USED,
};
struct Info{
bool operator<(const struct Info &arg) c... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#in... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include<bits/stdc++.h>
using namespace std;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
//INSERT ABOVE HERE
signed main(){
Int n;
cin>>n;
vector<Int> rs(n),as(n),ws(n),ts(n);
... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include<bits/stdc++.h>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef long double D;
typedef complex<D> P;
#define F first
#define S second
const ll MOD=1000000007;
//const ll MOD=998244353;
template<typename T,typename U>istream & operator >> (istr... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
#define rep(i, n) for(ll (i) = 0; (i) < (n); (i)++)
#define rep1(i, n) for(ll (i) = 1; (i) <= (n); (i)++)
#define rrep(i, n) for(ll (i) = (n) - 1; (i) >= 0; (i)--)
#define rrep1(i, n) for(ll (i) = (n); (i) >= 1; (i)--)
const ll ... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cassert>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <functional>
#include <iostream>
#include <map>
#include <set>
#include <cassert>
using namespace std;
typedef ... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include <iostream>
#include <string>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#define llint long long
#define inf 1000000000000000000
using namespace std;
llint n;
llint dp[2005][4005][2];
llint r[2005], a[2005], w[2005], t[2005];
llint order[2005];
bool comp(llint i, llint j)
{
return r[i]+t[i] ... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include<bits/stdc++.h>
using namespace std;
#define N 2000
int r[N], a[N], w[N], t[N];
int dp[N+1][N+1][2];
int main(){
int n;
cin >> n;
for(int i = 0; i < n; i++){
cin >> r[i] >> a[i] >> w[i] >> t[i];
}
vector<int> order(n);
iota(order.begin(), order.end(), 0);
sort(order.begin(),order.end(),[... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include <bits/stdc++.h>
#define INF 1e+18
#define int long long
using namespace std;
template<class T> T &chmin(T &a,const T &b) { return a = min(a,b); }
template<class T> T &chmax(T &a,const T &b) { return a = max(a,b); }
struct song{
int r,a,w,t;
};
int dp[2010][4050][2] = {};
signed main(){
int n;
cin >> n;
... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include <bits/stdc++.h>
using namespace std;
using lint = long long;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;
int main() {
cin.tie(nullptr); ios::sync_with_stdio(false);
int n; cin >> n;
V<> r(n), a(n), w(n), t(n); for (int i = 0; i < n; ++i) cin >> r[i] >> a[i]... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <cstring>
#include <queue>
#include <stack>
#include <math.h>
#include <iterator>
#include <vector>
#include <string>
#include <set>
#include <math.h>
#include <iostream>
#include <random>
#include... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
#include<map>
#include<queue>
#include<deque>
#include<iomanip>
#include<tuple>
using namespace std;
typedef long long int LL;
typedef pair<int,int> P;
typedef pair<LL,int> LP;
const int INF=1<<30... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include <bits/stdc++.h>
#define inf (long long)(1e9)
using namespace std;
struct data {
long long r, a, w, t;
};
bool asc(const data &l, const data &r) {
return l.t + l.r > r.t + r.r;
}
long long n;
vector<data> v;
long long dp[2005][4005][2] = {0};
long long solve();
int main() {
cin >> n;
v.resize(n);
... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
struct K {
int R, A, W, T;
bool operator<(const K &right)const {
return ... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
constexpr int INF = INT_MAX;
void chmax(ll &a, ll b){
a = max(a, b);
}
ll dp[2005][4005][2];
int main(){
int n;
cin >> n;
vector<P> ls;
vector<int> id(n), r(n), a(n), w(n), t(n);
for(int i=0;i<n;i+... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include<bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;
template<typename A,typename B>inline ... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include<iostream>
#include<vector>
#include<algorithm>
#include<utility>
using namespace std;
#define rep(i,n) for(int i = 0; i < (n); i++)
typedef pair<int,int> pii;
#define INF 1e9
struct data{
int r,a,w,t;
bool operator<(const data &another) const
{
return r+t < another.r + another.t;
};
};
... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
#define pb push_back
#define mp make_pair
#define eps 1e-9
#define INF 1000000000
#define LLINF 1000000000000000ll
#define sz(x) ((int)(x).size())
#define fi first
#define sec sec... |
Problem: problem
AOR Ika is a monster buster.
One day, as I was walking down the road, I met a sleeping monster.
AOR Ika, who has a strong fighting spirit, decides to hit the monster with a wake-up blow. However, the current attack power of AOR Ika is $ 0 $, and she cannot make a decent attack as it is.
AOR Ika-chan... | #include <bits/stdc++.h>
#define GET_MACRO(_1,_2,_3,_4,_5,_6,_7,_8,NAME,...) NAME
#define pr(...) cerr<< GET_MACRO(__VA_ARGS__,pr8,pr7,pr6,pr5,pr4,pr3,pr2,pr1)(__VA_ARGS__) <<endl
#define pr1(a) (#a)<<"="<<(a)<<" "
#define pr2(a,b) pr1(a)<<pr1(b)
#define pr3(a,b,c) pr1(a)<<pr2(b,c)
#define pr4(a,b,c,d) pr1(a)<<pr3(b,c,... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using vl = vector<ll>;
template<class T> using vc = vector<T>;
template<class T> using vvc = vector<vector<T>>;
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define rep(i, n) for (ll i = 0; i < (n); i++)
#defin... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #include<bits/stdc++.h>
using namespace std;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
//INSERT ABOVE HERE
signed main(){
Int n,l;
cin>>n>>l;
vector<Int> v(n),w(n),m(n);
for... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | import sys
from collections import deque
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
N, W = map(int, readline().split())
vs = [0]*N; ws = [0]*N; ms = [0]*N
for i in range(N):
vs[i], ws[i], ms[i] = map(int, readline().split())
V0 = max(vs)
V = sum(v * min(V0, m) for ... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #include<bits/stdc++.h>//c++14
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> mp;
ll inf = 1e15;
ll x = 25;
ll N,W;
vector<ll> dp,v,w,m;
ll solve(){//O(NWlogm)
vector<ll> dp(N*x*50+1,inf);
dp[0] = 0;
for(ll i=0;i<N;i++){
ll num = min(m[i],x) ;
m[i] -= num;
for(ll j=1;num > 0; j<<=1 ... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
struct Item {
ll v, w, m;
Item(){}
explicit Item(ll v, ll w, ll m):v(v), w(w), m(m){}
bool operator < (const Item& item) const {
return v*item.w < item.v*w;
}
};
constexpr ll MAX_N = 50;
constexpr ll MAX_V = 50;
constexpr ll LIMIT = M... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | // includes
#include <bits/stdc++.h>
// macros
#define ll long long int
#define pb emplace_back
#define mk make_pair
#define pq priority_queue
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #include <iostream>
#include <limits>
#include <vector>
#include <algorithm>
using ll = long long;
const ll INF = std::numeric_limits<ll>::max() / 2;
const ll VMAX = 50;
const ll NMAX = 50;
const ll S = VMAX * NMAX * VMAX;
// ナップザックDPにおける価値の最大値
struct Goods {
ll v, m, w;
};
int main() {
ll N, W;
std::ci... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
typedef unsigned long long ull;
typedef struct Item {
ull v;
ull w;
Item(ull v, ull w) : v(v), w(w) {};
} Item;
bool sort_by_cost(const Item& lhs, const Item& rhs) {
if ((__int128)lhs.v*rhs.w != (__int128)rhs.v*... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #!/usr/bin/env python3
# DPL_1_I: Combinatorial - Knapsack Problem with Limitations II
from heapq import heappush, heappop
def max_value(t, xs):
def _cost(weight, value, i):
for j in range(i, n):
v, w = xs[j]
if weight + w > t:
return (-value, -value - (v/w)*(t-wei... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #include <algorithm>
#include <cstring>
#include <deque>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <vector>
using namespace std;
using ll = long long;
using P = pair<ll, ll>; // (v, w)
#define fst first
#define snd second
vector<P> ps;
pair<l... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define all(x) (x).begin(),(x).end()
const int mod=1000000007,MAX=50001;
const ll INF=1LL<<60;
int main(){
int N;
ll W;cin>>N>>W;
ll ans=0;
vector<ll> v(N),w(N),m(N);
for(int i=0;i<N;i++){
cin>>v[i]>>w[i]>>m[i];
... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | from typing import Tuple, List
from heapq import heappush, heappop
def max_value(max_weight: int, items: List[Tuple[int, int]]) -> float:
def _cost(weight: int, value: int, i: int) -> Tuple[int, float]:
# Use value as negative cost
for j in range(i, n):
v, w = items[j]
if (... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | def main():
from collections import deque
from functools import cmp_to_key
N, W, *A = map(int, open(0).read().split())
A = [list(a) for a in zip(*[iter(A)]*3)]
Vmax = max(v for v, w, m in A)
Vsum = 0
B = []
for i, (v, w, m) in enumerate(A):
m_upper = min(m, Vmax-1)
... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
vector<pll> items;
ll best;
bool cmp(const pll &a,const pll &b)
{
ll x=a.first*b.second,y=a.second*b.first;
if(x!=y) return x>y;
else return a.first>b.first;
}
pair<ll,bool> estimate(int i,ll w)
{
ll r... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>
using namespace std;
#define rep(i,n) for(i = 0;i < n;++i)
#define all(v) v.begin(), v.end()
using ll = long long;
int main()
{
ll i,j;
ll n, W;
cin >> n >> W;
vector<ll> v(n);
vector<ll> w(n);
vector<ll> m(n);
f... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #!python3
iim = lambda: map(int, input().rstrip().split())
from heapq import heappush, heappop
def resolve():
N, W = iim()
S = [list(iim()) for i in range(N)]
def f1(v, w, m):
mm = []
i = 1
while i <= m:
yield i
m -= i
i <<= 1
if m:
... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #include <bits/stdc++.h>
using namespace std;
typedef pair< long long, long long > P;
bool sort_cost( const P &l, const P &r )
{
if( (__int128)l.second * r.first != (__int128)r.second * l.first ) {
return (__int128)l.second * r.first > (__int128)r.second * l.first;
}
else {
return l.second > r.second;
}
}
l... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #include <bits/stdc++.h>
#include <cmath>
using namespace std;
typedef long long ll;
//typedef unsigned long long ll;
#define rep(i, n) for (int i = 0; i < (n); ++i)
//#define rep(i, n) for (ll i = 0; i < (n); ++i)
//#define sz(x) ll(x.size())
//typedef pair<ll, int> P;
typedef pair<ll, ll> P;
//const double INF = 1e... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | // includes
#include <bits/stdc++.h>
// macros
#define ll long long int
#define pb emplace_back
#define mk make_pair
#define pq priority_queue
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #include <bits/stdc++.h>
#define show(x) std::cerr << #x << " = " << (x) << std::endl
using ll = long long;
template <typename T>
constexpr T INF() { return std::numeric_limits<T>::max() / 16; }
template <typename T, typename A>
std::ostream& operator<<(std::ostream& os, const std::vector<T, A>& v)
{
os << "[";
... |
Problem: You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsac... | #include <iostream>
#include <vector>
#include <deque>
#include <algorithm>
using namespace std;
class Item {
public:
long long v, w, m;
explicit Item(long long v, long long w, long long m) : v(v), w(w), m(m) {}
bool operator < (const Item& it) const { return v * it.w < it.v * w; }
};
int main(){
int... |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | #include <bits/stdc++.h>
using namespace std;
int main()
{
long long n,m;
cin>>n;
m=sqrt(n);
cout<<n<<":";
for(int i=2;i<=n;i++)
{
while(n%i==0){
cout<<" "<<i;
n/=i;
}
if(n==1)
break;
if(i>=sqrt(n))
{
cout<<" "<<n;
brea... |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
int tmp = 2;
queue<int> que;
cin >> n;
int ans = n;
while(n > 1 && tmp*tmp <= n){
if(n % tmp == 0){
que.push(tmp);
n /= tmp;
}else{
tmp++;
}
}
if(n > 1){
que.push(n);
}
cout << ans << ":";
while... |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... |
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class Main {
void run() {
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
Map<Long, Long> map = primeFactorization(n);
System.out.print(n + ":");
map.ent... |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | def primefactor(x):
fac=[]
for i in xrange(2,int(x**0.5)+1):
while x%i==0:
fac+=[i]
x=x/i
if x!=1:
fac+=[x]
return fac
n=input()
ans=primefactor(n)
print(str(n)+":"+" "+" ".join(map(str,ans))) |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | //
#include<iostream>
using namespace std;
int main() {
int n;
cin >> n;
int x=n;
cout << n << ":";
for(int i=2; i*i<=x; i++) {
while(n%i==0) {
cout << " " << i;
n/=i;
}
}
if(n!=1)
cout << " " << n;
cout << endl;
return 0;
}
|
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | #include <iostream>
#include <cmath>
#include <sstream>
using namespace std;
int main()
{
int n;
cin >> n;
int limit = sqrt(n);
stringstream ss;
ss << n << ":" ;
for (int i = 2; i <= sqrt(n); i++) {
//int count;
while (n % i == 0) {
n /= i;
ss << " " << i;
//count ++;
}
}
... |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | import math
n=int(input())
factors=[]
n1=n
p=2
while p<=math.sqrt(n):
if n%p==0:
n//=p
factors.append(str(p))
else:
p+=1
if n!=1:
factors.append(str(n))
print(str(n1)+": "+" ".join(factors))
|
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | import java.util.Scanner;
//?´??????°????§£?????????????????°??????
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long n = Long.parseLong(sc.nextLine());
System.out.print(n + ":");
for (int i = 2; n >= i * i;) {
if (n % i == 0) {
n = n / i;
Syste... |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
ll n;
cin >> n;
vector<ll> ans;
cout << n << ":";
for(ll i = 2; i*i <= n; i++){
while(n % i == 0){
ans.emplace_back(i);
n /= i;
}
}
if(n != 1) ans.emplace_back(n);
... |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | m=input();n=int(m);s=n**.5;a=[];i=6
while n%2==0:a+=[2];n//=2
while n%3==0:a+=[3];n//=3
while i<s:
if n%(i-1)==0:a+=[i-1];n//=i-1
elif n%(i+1)==0:a+=[i+1];n//=i+1
else:i+=6
if n>1:a+=[n]
elif n==int(m):a+=[m]
print(m+':',*a)
|
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | #include<iostream>
#include<stdio.h>
#include<vector>
#include<math.h>
#include<algorithm>
using namespace std;
int main()
{
int n;
cin >> n;
cout << n << ":";
int ren = n;
for (int i = 1; i <= sqrt(n); i++)
{
while (ren != 1)
{
if (ren % (i + 1) == 0)
{
cout << " " <<i + 1;
ren = ren / (i +... |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | #include <bits/stdc++.h>
using namespace std;
signed main(void)
{
long long num; cin >> num;
cout << num << ":";
for (int i = 2; i*i <= num; ++i)
while (num % i == 0)cout << " " << i, num /= i;
if (num != 1)cout << " " << num; cout << endl;
} |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | n=int(input())
s=str(n)+":"
b=True
d=n
i=2
while b:
if d%i==0:
d/=i
s+=' '+str(i)
i-=1
b=i<d**0.5
i+=1
if d!=1:s+=' '+str(int(d))
print(s) |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | import java.util.ArrayList;
import java.util.Scanner;
public class Main {
private static int nowSosuu = 1;
private static ArrayList<Integer> array = new ArrayList<Integer>();
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
System.out.printf(n + ":");
... |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | #include<iostream>
#include<cmath>
using namespace std;
int main(){
int n;
cin>>n;
cout<<n<<":";
for(int i=2;i<=sqrt((double)n);){
if(n%i==0){
cout<<" "<<i;
n/=i;
}
else i++;
}
if(n!=0) cout<<" "<<n;
cout<<endl;
} |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | #include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
int main(){
long long n,i;
cin>>n;
cout<<n<<":";
i=2;
while(n>1 && i*i<=n){
if(n%i==0){
cout<<" "<<i;
n/=i;
}
else i+=1;
}
if(n!=1)cout<<" "<<n;
cout<<endl;
return 0;
} |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | import java.util.Scanner;
import java.util.*;
public class Main{
public static void main(String[] args){
ArrayList<Integer> list =new ArrayList<Integer>();
Scanner sc = new Scanner(System.in);
int a= sc.nextInt();
int b=a;
for(int i=2;i<b/2;i++){
while(a%i==0){
... |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const ll mod = 1000000007;
const int inf = 1e9;
const long long INF = 1LL << 60; // INFの値は1152921504606846976
int main()
{
int n;
cin >> n;
cout << n << ':';
for(int i = 2; i * i <= n; i++){
while(... |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | #include <iostream>
#include <stdio.h>
using namespace std;
bool IsPrime(int n){
for(int i=2;i*i<=n;i++){
if(n%i==0) return false;
}
return true;
}
int main() {
int n;
cin >> n;
cout << n <<":";
if(IsPrime(n)) cout << " " << n;
else{
while(IsPrime(n)!=true){
for(int i=2;i<=n&&n!=1;i++){
whil... |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | p=lambda x:print(x,end=' ')
n=input();p(n+':')
n=int(n);s=n**.5
while n%2==0 and n>3:p(2);n//=2
d=3
while s>d and n>d:
if n%d>0:d+=2
else:p(d);n//=d
print(n)
|
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | #include<iostream>
using namespace std;
int main()
{
int n;
cin>>n;
int save=n;
cout<<n<<":";
for(int i=2;i*i<=save;i++)
if(n%i==0)
{
while(n%i==0){
n/=i;
cout<<" "<<i;
}
}
if(n>1)
cout<<" "<<n;
cout<<endl;
return 0;
} |
Problem: Factorize a given integer n.
Constraints
* 2 ≤ n ≤ 109
Input
n
An integer n is given in a line.
Output
Print the given integer n and :. Then, print prime factors in ascending order. If n is divisible by a prime factor several times, the prime factor should be printed according to the number of times. ... | def prime_factorize(n):
a=[]
while n%2==0:
a.append(2)
n //=2
f=3
while f**2<=n:
if n%f==0:
a.append(f)
n //=f
else:f +=2
if n!=1:a.append(n)
return a
n=int(input())
A=prime_factorize(n)
print("{}:".format(n),*A)
|
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