stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
// Author: Amey Bhavsar

// IDE: Geany/Ubuntu

// Platform: *



#include<bits/stdc++.h>

using namespace std;



typedef long long ll;



#define rep(i,a,b) for(auto i=a;i<b;i++)

#define repD(i,a,b) for(auto i=a;i>=b;i--)

#define pb push_back

#define mp make_pair

#define ff first

#define ss second



// to be used with algorithms that processes a container Eg: find(all(c),42)

#define all(x) (x).begin(), (x).end() //Forward traversal

#define rall(x) (x).rbegin, (x).rend() //reverse traversal

// traversal function to avoid long template definition

#define tr(c,i) for(__typeof__((c)).begin() i = (c).begin(); i != (c).end(); i++)

//find version works for all containers. This is present in std namespace.

#define cpresent(c,x) (find(all(c),x) != (c).end())

// ll fe (ll x, ll y) {

	// ll ans = 1;

	// while (y>0) {

		// if (y&1) {

			// ans = (x*ans) % MOD;

		// }

		// y /= 2;

		// x = (x*x) % MOD;

	// }

	// return (ans % MOD);

// }



// ll modmul(ll a, ll b) {

    // return ((a%MOD) * (b%MOD)) % MOD;

// }



// ll modadd(ll a , ll b){

    // return((a%MOD)+(b%MOD)+MOD)%MOD;

// }



// ll modsub(ll a , ll b){

    // return((a%MOD) - (b%MOD) + MOD)%MOD;

// }

#define MOD 1000000007

// int n;

// vector<int> a;

// void solution() {

	// cin >> n;

	// a.resize(n);

	// for (auto& e:a) {

		// cin >> e;

	// }

// }



int main() {

	ios_base::sync_with_stdio(false);cin.tie(NULL);

	int a,b;

	cin >> a >> b;

	if (a >= 10) {

		cout << b;

	} else {

		cout << 100*(10-a) + b;

	}

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.326150476384193x):
#include<stdio.h>



int main(){

int a,b;



scanf("%d%d", &a, &b);

if(a<10)printf("%d\n", (100*(10-a)+b));

else printf("%d\n", b);



	return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

#define FLASH ios_base::sync_with_stdio(0);

#define ll long long

#define debt(x,y)cout<<"#x = "<<(x)<<" and "<<"#y = "<<(y)<<endl;

#define deb(x)cout<<"#x = "<<(x)<<endl;

#define pb push_back

#define mp make_pair

#define all(x) (x).begin(), (x).end()

#define endl "\n"

#define arr(a,n) for(ll i=1;i<=n;i++) cout<<a[i]<<" "; cout << "\n";

#define vecc(a,n) for(ll i=0;i<n;i++) cout<<a[i]<<" "; cout << "\n";





using namespace std;



ll n,r;



int main(){

FLASH;

cin>>n>>r;

if(n<10)cout<<r+(100*(10-n))<<endl;

else cout<<r<<endl;













return 0;

} 


// Faster C++ solution for the same problem (measured speedup: 5.336874561116191x):
#include<stdio.h>

	int main(){

		int N,R,hasil;

		

		scanf ("%d %d",&N,&R);

		if (N<10){

			hasil = 100*(10-N);

			hasil = R + hasil;

		}

		else {

			hasil= R;

		}

		printf ("%d", hasil);

	}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



int a,b;

int main() {

cin>>a>>b;

if(a<10){

	cout<<b+(100*(10-a));

}else{

	cout<<b;

}

}

// Faster C++ solution for the same problem (measured speedup: 5.257060594831893x):
#include<stdio.h>

	int main(){

		int N,R,hasil;

		

		scanf ("%d %d",&N,&R);

		if (N<10){

			hasil = 100*(10-N);

			hasil = R + hasil;

		}

		else {

			hasil= R;

		}

		printf ("%d", hasil);

	}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main ()

{

    int N,R,S,A;

    cin>>N>>R;

    if(N<10)

    {

    S=100*(10-N);

    A=R+S;

    cout<<A<<endl;

    }

    else

        cout<<R<<endl;

    return 0;



}

// Faster C++ solution for the same problem (measured speedup: 5.266404210221221x):
#include<stdio.h>

	int main(){

		int N,R,hasil;

		

		scanf ("%d %d",&N,&R);

		if (N<10){

			hasil = 100*(10-N);

			hasil = R + hasil;

		}

		else {

			hasil= R;

		}

		printf ("%d", hasil);

	}
// Slower C++ solution:
#include <iostream>

using namespace std;

//zakha



int N,R;



int main () {

    cin >> N >> R;

    if (N >= 10) {

        cout << R << endl;

    } else {

        cout << R + 100*(10-N) << endl;

    

    }

}




// Faster C++ solution for the same problem (measured speedup: 5.266182838995032x):
#include<stdio.h>

	int main(){

		int N,R,hasil;

		

		scanf ("%d %d",&N,&R);

		if (N<10){

			hasil = 100*(10-N);

			hasil = R + hasil;

		}

		else {

			hasil= R;

		}

		printf ("%d", hasil);

	}
// Slower C++ solution:
#include <iostream>

using namespace std;



// Caleb Faith Adnaka XI105

int main() {



    int N,R;

    cin >> N;

    cin >> R;



    if (N >= 10){

        cout << R << "\n";

    }

    else{

        cout << R + (100 * (10-N)) << "\n";

    }



}

// Faster C++ solution for the same problem (measured speedup: 5.268796112255285x):
#include<stdio.h>

	int main(){

		int N,R,hasil;

		

		scanf ("%d %d",&N,&R);

		if (N<10){

			hasil = 100*(10-N);

			hasil = R + hasil;

		}

		else {

			hasil= R;

		}

		printf ("%d", hasil);

	}
// Slower C++ solution:
// Author: Amey Bhavsar

// IDE: Geany/Ubuntu

// Platform: *



#include<bits/stdc++.h>

using namespace std;



typedef long long ll;



#define rep(i,a,b) for(auto i=a;i<b;i++)

#define repD(i,a,b) for(auto i=a;i>=b;i--)

#define pb push_back

#define mp make_pair

#define ff first

#define ss second



// to be used with algorithms that processes a container Eg: find(all(c),42)

#define all(x) (x).begin(), (x).end() //Forward traversal

#define rall(x) (x).rbegin, (x).rend() //reverse traversal

// traversal function to avoid long template definition

#define tr(c,i) for(__typeof__((c)).begin() i = (c).begin(); i != (c).end(); i++)

//find version works for all containers. This is present in std namespace.

#define cpresent(c,x) (find(all(c),x) != (c).end())

// ll fe (ll x, ll y) {

	// ll ans = 1;

	// while (y>0) {

		// if (y&1) {

			// ans = (x*ans) % MOD;

		// }

		// y /= 2;

		// x = (x*x) % MOD;

	// }

	// return (ans % MOD);

// }



// ll modmul(ll a, ll b) {

    // return ((a%MOD) * (b%MOD)) % MOD;

// }



// ll modadd(ll a , ll b){

    // return((a%MOD)+(b%MOD)+MOD)%MOD;

// }



// ll modsub(ll a , ll b){

    // return((a%MOD) - (b%MOD) + MOD)%MOD;

// }

#define MOD 1000000007

// int n;

// vector<int> a;

// void solution() {

	// cin >> n;

	// a.resize(n);

	// for (auto& e:a) {

		// cin >> e;

	// }

// }



int main() {

	ios_base::sync_with_stdio(false);cin.tie(NULL);

	int a,b;

	cin >> a >> b;

	if (a >= 10) {

		cout << b;

	} else {

		cout << 100*(10-a) + b;

	}

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.3319616692430225x):
#include<stdio.h>

	int main(){

		int N,R,hasil;

		

		scanf ("%d %d",&N,&R);

		if (N<10){

			hasil = 100*(10-N);

			hasil = R + hasil;

		}

		else {

			hasil= R;

		}

		printf ("%d", hasil);

	}
// Slower C++ solution:
#include <bits/stdc++.h>

#define FLASH ios_base::sync_with_stdio(0);

#define ll long long

#define debt(x,y)cout<<"#x = "<<(x)<<" and "<<"#y = "<<(y)<<endl;

#define deb(x)cout<<"#x = "<<(x)<<endl;

#define pb push_back

#define mp make_pair

#define all(x) (x).begin(), (x).end()

#define endl "\n"

#define arr(a,n) for(ll i=1;i<=n;i++) cout<<a[i]<<" "; cout << "\n";

#define vecc(a,n) for(ll i=0;i<n;i++) cout<<a[i]<<" "; cout << "\n";





using namespace std;



ll n,r;



int main(){

FLASH;

cin>>n>>r;

if(n<10)cout<<r+(100*(10-n))<<endl;

else cout<<r<<endl;













return 0;

} 


// Faster C++ solution for the same problem (measured speedup: 5.339305338338455x):
#include <stdio.h>



int main(){

	int N, R;

	scanf("%d %d", &N, &R);

	

	if (N<= 10){

		printf("%d", (100*(10-N))+R);

	}else if (N>10){

		printf("%d", R);

	}

	

	

	return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



int a,b;

int main() {

cin>>a>>b;

if(a<10){

	cout<<b+(100*(10-a));

}else{

	cout<<b;

}

}

// Faster C++ solution for the same problem (measured speedup: 5.259455019322041x):
#include <stdio.h>



int main(){

	int N, R;

	scanf("%d %d", &N, &R);

	

	if (N<= 10){

		printf("%d", (100*(10-N))+R);

	}else if (N>10){

		printf("%d", R);

	}

	

	

	return 0;

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main ()

{

    int N,R,S,A;

    cin>>N>>R;

    if(N<10)

    {

    S=100*(10-N);

    A=R+S;

    cout<<A<<endl;

    }

    else

        cout<<R<<endl;

    return 0;



}

// Faster C++ solution for the same problem (measured speedup: 5.268802890432053x):
#include <stdio.h>



int main(){

	int N, R;

	scanf("%d %d", &N, &R);

	

	if (N<= 10){

		printf("%d", (100*(10-N))+R);

	}else if (N>10){

		printf("%d", R);

	}

	

	

	return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;

//zakha



int N,R;



int main () {

    cin >> N >> R;

    if (N >= 10) {

        cout << R << endl;

    } else {

        cout << R + 100*(10-N) << endl;

    

    }

}




// Faster C++ solution for the same problem (measured speedup: 5.268581418378288x):
#include <stdio.h>



int main(){

	int N, R;

	scanf("%d %d", &N, &R);

	

	if (N<= 10){

		printf("%d", (100*(10-N))+R);

	}else if (N>10){

		printf("%d", R);

	}

	

	

	return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;



// Caleb Faith Adnaka XI105

int main() {



    int N,R;

    cin >> N;

    cin >> R;



    if (N >= 10){

        cout << R << "\n";

    }

    else{

        cout << R + (100 * (10-N)) << "\n";

    }



}

// Faster C++ solution for the same problem (measured speedup: 5.271195881901688x):
#include <stdio.h>



int main(){

	int N, R;

	scanf("%d %d", &N, &R);

	

	if (N<= 10){

		printf("%d", (100*(10-N))+R);

	}else if (N>10){

		printf("%d", R);

	}

	

	

	return 0;

}
// Slower C++ solution:
// Author: Amey Bhavsar

// IDE: Geany/Ubuntu

// Platform: *



#include<bits/stdc++.h>

using namespace std;



typedef long long ll;



#define rep(i,a,b) for(auto i=a;i<b;i++)

#define repD(i,a,b) for(auto i=a;i>=b;i--)

#define pb push_back

#define mp make_pair

#define ff first

#define ss second



// to be used with algorithms that processes a container Eg: find(all(c),42)

#define all(x) (x).begin(), (x).end() //Forward traversal

#define rall(x) (x).rbegin, (x).rend() //reverse traversal

// traversal function to avoid long template definition

#define tr(c,i) for(__typeof__((c)).begin() i = (c).begin(); i != (c).end(); i++)

//find version works for all containers. This is present in std namespace.

#define cpresent(c,x) (find(all(c),x) != (c).end())

// ll fe (ll x, ll y) {

	// ll ans = 1;

	// while (y>0) {

		// if (y&1) {

			// ans = (x*ans) % MOD;

		// }

		// y /= 2;

		// x = (x*x) % MOD;

	// }

	// return (ans % MOD);

// }



// ll modmul(ll a, ll b) {

    // return ((a%MOD) * (b%MOD)) % MOD;

// }



// ll modadd(ll a , ll b){

    // return((a%MOD)+(b%MOD)+MOD)%MOD;

// }



// ll modsub(ll a , ll b){

    // return((a%MOD) - (b%MOD) + MOD)%MOD;

// }

#define MOD 1000000007

// int n;

// vector<int> a;

// void solution() {

	// cin >> n;

	// a.resize(n);

	// for (auto& e:a) {

		// cin >> e;

	// }

// }



int main() {

	ios_base::sync_with_stdio(false);cin.tie(NULL);

	int a,b;

	cin >> a >> b;

	if (a >= 10) {

		cout << b;

	} else {

		cout << 100*(10-a) + b;

	}

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.334390208798743x):
#include <stdio.h>



int main(){

	int N, R;

	scanf("%d %d", &N, &R);

	

	if (N<= 10){

		printf("%d", (100*(10-N))+R);

	}else if (N>10){

		printf("%d", R);

	}

	

	

	return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

#define FLASH ios_base::sync_with_stdio(0);

#define ll long long

#define debt(x,y)cout<<"#x = "<<(x)<<" and "<<"#y = "<<(y)<<endl;

#define deb(x)cout<<"#x = "<<(x)<<endl;

#define pb push_back

#define mp make_pair

#define all(x) (x).begin(), (x).end()

#define endl "\n"

#define arr(a,n) for(ll i=1;i<=n;i++) cout<<a[i]<<" "; cout << "\n";

#define vecc(a,n) for(ll i=0;i<n;i++) cout<<a[i]<<" "; cout << "\n";





using namespace std;



ll n,r;



int main(){

FLASH;

cin>>n>>r;

if(n<10)cout<<r+(100*(10-n))<<endl;

else cout<<r<<endl;













return 0;

} 


// Faster C++ solution for the same problem (measured speedup: 5.330470963770527x):
#include <stdio.h>



int main(){

	int N,R;

	

	scanf("%d %d", &N, &R);

	

	if(N>=10){

		printf("%d\n", R);

	}

	else{

		printf("%d\n", R+100*(10-N));

	}

	return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



int a,b;

int main() {

cin>>a>>b;

if(a<10){

	cout<<b+(100*(10-a));

}else{

	cout<<b;

}

}

// Faster C++ solution for the same problem (measured speedup: 5.250752764493078x):
#include <stdio.h>



int main(){

	int N,R;

	

	scanf("%d %d", &N, &R);

	

	if(N>=10){

		printf("%d\n", R);

	}

	else{

		printf("%d\n", R+100*(10-N));

	}

	return 0;

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main ()

{

    int N,R,S,A;

    cin>>N>>R;

    if(N<10)

    {

    S=100*(10-N);

    A=R+S;

    cout<<A<<endl;

    }

    else

        cout<<R<<endl;

    return 0;



}

// Faster C++ solution for the same problem (measured speedup: 5.260085168685662x):
#include <stdio.h>



int main(){

	int N,R;

	

	scanf("%d %d", &N, &R);

	

	if(N>=10){

		printf("%d\n", R);

	}

	else{

		printf("%d\n", R+100*(10-N));

	}

	return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;

//zakha



int N,R;



int main () {

    cin >> N >> R;

    if (N >= 10) {

        cout << R << endl;

    } else {

        cout << R + 100*(10-N) << endl;

    

    }

}




// Faster C++ solution for the same problem (measured speedup: 5.259864063077895x):
#include <stdio.h>



int main(){

	int N,R;

	

	scanf("%d %d", &N, &R);

	

	if(N>=10){

		printf("%d\n", R);

	}

	else{

		printf("%d\n", R+100*(10-N));

	}

	return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;



// Caleb Faith Adnaka XI105

int main() {



    int N,R;

    cin >> N;

    cin >> R;



    if (N >= 10){

        cout << R << "\n";

    }

    else{

        cout << R + (100 * (10-N)) << "\n";

    }



}

// Faster C++ solution for the same problem (measured speedup: 5.262474200729558x):
#include <stdio.h>



int main(){

	int N,R;

	

	scanf("%d %d", &N, &R);

	

	if(N>=10){

		printf("%d\n", R);

	}

	else{

		printf("%d\n", R+100*(10-N));

	}

	return 0;

}
// Slower C++ solution:
// Author: Amey Bhavsar

// IDE: Geany/Ubuntu

// Platform: *



#include<bits/stdc++.h>

using namespace std;



typedef long long ll;



#define rep(i,a,b) for(auto i=a;i<b;i++)

#define repD(i,a,b) for(auto i=a;i>=b;i--)

#define pb push_back

#define mp make_pair

#define ff first

#define ss second



// to be used with algorithms that processes a container Eg: find(all(c),42)

#define all(x) (x).begin(), (x).end() //Forward traversal

#define rall(x) (x).rbegin, (x).rend() //reverse traversal

// traversal function to avoid long template definition

#define tr(c,i) for(__typeof__((c)).begin() i = (c).begin(); i != (c).end(); i++)

//find version works for all containers. This is present in std namespace.

#define cpresent(c,x) (find(all(c),x) != (c).end())

// ll fe (ll x, ll y) {

	// ll ans = 1;

	// while (y>0) {

		// if (y&1) {

			// ans = (x*ans) % MOD;

		// }

		// y /= 2;

		// x = (x*x) % MOD;

	// }

	// return (ans % MOD);

// }



// ll modmul(ll a, ll b) {

    // return ((a%MOD) * (b%MOD)) % MOD;

// }



// ll modadd(ll a , ll b){

    // return((a%MOD)+(b%MOD)+MOD)%MOD;

// }



// ll modsub(ll a , ll b){

    // return((a%MOD) - (b%MOD) + MOD)%MOD;

// }

#define MOD 1000000007

// int n;

// vector<int> a;

// void solution() {

	// cin >> n;

	// a.resize(n);

	// for (auto& e:a) {

		// cin >> e;

	// }

// }



int main() {

	ios_base::sync_with_stdio(false);cin.tie(NULL);

	int a,b;

	cin >> a >> b;

	if (a >= 10) {

		cout << b;

	} else {

		cout << 100*(10-a) + b;

	}

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.325563966767287x):
#include <stdio.h>



int main(){

	int N,R;

	

	scanf("%d %d", &N, &R);

	

	if(N>=10){

		printf("%d\n", R);

	}

	else{

		printf("%d\n", R+100*(10-N));

	}

	return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

#define FLASH ios_base::sync_with_stdio(0);

#define ll long long

#define debt(x,y)cout<<"#x = "<<(x)<<" and "<<"#y = "<<(y)<<endl;

#define deb(x)cout<<"#x = "<<(x)<<endl;

#define pb push_back

#define mp make_pair

#define all(x) (x).begin(), (x).end()

#define endl "\n"

#define arr(a,n) for(ll i=1;i<=n;i++) cout<<a[i]<<" "; cout << "\n";

#define vecc(a,n) for(ll i=0;i<n;i++) cout<<a[i]<<" "; cout << "\n";





using namespace std;



ll n,r;



int main(){

FLASH;

cin>>n>>r;

if(n<10)cout<<r+(100*(10-n))<<endl;

else cout<<r<<endl;













return 0;

} 


// Faster C++ solution for the same problem (measured speedup: 5.339979950496341x):
#include <stdio.h>



int main()

{

	int c, dr; 

	scanf("%d %d", &c, &dr);

	

	if(c >=10)

{

	printf("%d" , dr);

}

    else

{

	printf("%d", dr+100*(10-c));

}

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



int a,b;

int main() {

cin>>a>>b;

if(a<10){

	cout<<b+(100*(10-a));

}else{

	cout<<b;

}

}

// Faster C++ solution for the same problem (measured speedup: 5.260119542527786x):
#include <stdio.h>



int main()

{

	int c, dr; 

	scanf("%d %d", &c, &dr);

	

	if(c >=10)

{

	printf("%d" , dr);

}

    else

{

	printf("%d", dr+100*(10-c));

}

    return 0;

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main ()

{

    int N,R,S,A;

    cin>>N>>R;

    if(N<10)

    {

    S=100*(10-N);

    A=R+S;

    cout<<A<<endl;

    }

    else

        cout<<R<<endl;

    return 0;



}

// Faster C++ solution for the same problem (measured speedup: 5.269468594725431x):
#include <stdio.h>



int main()

{

	int c, dr; 

	scanf("%d %d", &c, &dr);

	

	if(c >=10)

{

	printf("%d" , dr);

}

    else

{

	printf("%d", dr+100*(10-c));

}

    return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;

//zakha



int N,R;



int main () {

    cin >> N >> R;

    if (N >= 10) {

        cout << R << endl;

    } else {

        cout << R + 100*(10-N) << endl;

    

    }

}




// Faster C++ solution for the same problem (measured speedup: 5.269247094689048x):
#include <stdio.h>



int main()

{

	int c, dr; 

	scanf("%d %d", &c, &dr);

	

	if(c >=10)

{

	printf("%d" , dr);

}

    else

{

	printf("%d", dr+100*(10-c));

}

    return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;



// Caleb Faith Adnaka XI105

int main() {



    int N,R;

    cin >> N;

    cin >> R;



    if (N >= 10){

        cout << R << "\n";

    }

    else{

        cout << R + (100 * (10-N)) << "\n";

    }



}

// Faster C++ solution for the same problem (measured speedup: 5.271861888545473x):
#include <stdio.h>



int main()

{

	int c, dr; 

	scanf("%d %d", &c, &dr);

	

	if(c >=10)

{

	printf("%d" , dr);

}

    else

{

	printf("%d", dr+100*(10-c));

}

    return 0;

}
// Slower C++ solution:
// Author: Amey Bhavsar

// IDE: Geany/Ubuntu

// Platform: *



#include<bits/stdc++.h>

using namespace std;



typedef long long ll;



#define rep(i,a,b) for(auto i=a;i<b;i++)

#define repD(i,a,b) for(auto i=a;i>=b;i--)

#define pb push_back

#define mp make_pair

#define ff first

#define ss second



// to be used with algorithms that processes a container Eg: find(all(c),42)

#define all(x) (x).begin(), (x).end() //Forward traversal

#define rall(x) (x).rbegin, (x).rend() //reverse traversal

// traversal function to avoid long template definition

#define tr(c,i) for(__typeof__((c)).begin() i = (c).begin(); i != (c).end(); i++)

//find version works for all containers. This is present in std namespace.

#define cpresent(c,x) (find(all(c),x) != (c).end())

// ll fe (ll x, ll y) {

	// ll ans = 1;

	// while (y>0) {

		// if (y&1) {

			// ans = (x*ans) % MOD;

		// }

		// y /= 2;

		// x = (x*x) % MOD;

	// }

	// return (ans % MOD);

// }



// ll modmul(ll a, ll b) {

    // return ((a%MOD) * (b%MOD)) % MOD;

// }



// ll modadd(ll a , ll b){

    // return((a%MOD)+(b%MOD)+MOD)%MOD;

// }



// ll modsub(ll a , ll b){

    // return((a%MOD) - (b%MOD) + MOD)%MOD;

// }

#define MOD 1000000007

// int n;

// vector<int> a;

// void solution() {

	// cin >> n;

	// a.resize(n);

	// for (auto& e:a) {

		// cin >> e;

	// }

// }



int main() {

	ios_base::sync_with_stdio(false);cin.tie(NULL);

	int a,b;

	cin >> a >> b;

	if (a >= 10) {

		cout << b;

	} else {

		cout << 100*(10-a) + b;

	}

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.335064199938361x):
#include <stdio.h>



int main()

{

	int c, dr; 

	scanf("%d %d", &c, &dr);

	

	if(c >=10)

{

	printf("%d" , dr);

}

    else

{

	printf("%d", dr+100*(10-c));

}

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

#define FLASH ios_base::sync_with_stdio(0);

#define ll long long

#define debt(x,y)cout<<"#x = "<<(x)<<" and "<<"#y = "<<(y)<<endl;

#define deb(x)cout<<"#x = "<<(x)<<endl;

#define pb push_back

#define mp make_pair

#define all(x) (x).begin(), (x).end()

#define endl "\n"

#define arr(a,n) for(ll i=1;i<=n;i++) cout<<a[i]<<" "; cout << "\n";

#define vecc(a,n) for(ll i=0;i<n;i++) cout<<a[i]<<" "; cout << "\n";





using namespace std;



ll n,r;



int main(){

FLASH;

cin>>n>>r;

if(n<10)cout<<r+(100*(10-n))<<endl;

else cout<<r<<endl;













return 0;

} 


// Faster C++ solution for the same problem (measured speedup: 5.339083652080626x):
#include <stdio.h>



int main()

{

	int N, R;

	scanf("%d%d",&N,&R);

	if(R>4111)

	{

		printf("second value must between 0-4111");

	}

	else if(N>=10)

	{

		printf("%d", R);

	}

	else

	{

		printf("%d", R+100*(10-N));

	}

	return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



int a,b;

int main() {

cin>>a>>b;

if(a<10){

	cout<<b+(100*(10-a));

}else{

	cout<<b;

}

}

// Faster C++ solution for the same problem (measured speedup: 5.259236648424037x):
#include <stdio.h>



int main()

{

	int N, R;

	scanf("%d%d",&N,&R);

	if(R>4111)

	{

		printf("second value must between 0-4111");

	}

	else if(N>=10)

	{

		printf("%d", R);

	}

	else

	{

		printf("%d", R+100*(10-N));

	}

	return 0;

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main ()

{

    int N,R,S,A;

    cin>>N>>R;

    if(N<10)

    {

    S=100*(10-N);

    A=R+S;

    cout<<A<<endl;

    }

    else

        cout<<R<<endl;

    return 0;



}

// Faster C++ solution for the same problem (measured speedup: 5.268584131413417x):
#include <stdio.h>



int main()

{

	int N, R;

	scanf("%d%d",&N,&R);

	if(R>4111)

	{

		printf("second value must between 0-4111");

	}

	else if(N>=10)

	{

		printf("%d", R);

	}

	else

	{

		printf("%d", R+100*(10-N));

	}

	return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;

//zakha



int N,R;



int main () {

    cin >> N >> R;

    if (N >= 10) {

        cout << R << endl;

    } else {

        cout << R + 100*(10-N) << endl;

    

    }

}




// Faster C++ solution for the same problem (measured speedup: 5.2683626685551x):
#include <stdio.h>



int main()

{

	int N, R;

	scanf("%d%d",&N,&R);

	if(R>4111)

	{

		printf("second value must between 0-4111");

	}

	else if(N>=10)

	{

		printf("%d", R);

	}

	else

	{

		printf("%d", R+100*(10-N));

	}

	return 0;

}
// Slower C++ solution:
#include <iostream>

using namespace std;



// Caleb Faith Adnaka XI105

int main() {



    int N,R;

    cin >> N;

    cin >> R;



    if (N >= 10){

        cout << R << "\n";

    }

    else{

        cout << R + (100 * (10-N)) << "\n";

    }



}

// Faster C++ solution for the same problem (measured speedup: 5.270977023526808x):
#include <stdio.h>



int main()

{

	int N, R;

	scanf("%d%d",&N,&R);

	if(R>4111)

	{

		printf("second value must between 0-4111");

	}

	else if(N>=10)

	{

		printf("%d", R);

	}

	else

	{

		printf("%d", R+100*(10-N));

	}

	return 0;

}