stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include<iostream>

#include<cstring>

#include<cstdio>

using namespace std;

typedef long long ll;

const int maxn = 1000010;

const int inf = 0x3f3f3f3f;

ll dp[maxn];

ll n, W;

ll v[maxn];

ll w[maxn];

inline ll max(ll x, ll y) {

	return x > y ? x : y;

}

inline ll min(ll x, ll y) {

	return x < y ? x : y;

}

int main() {

	memset(dp, 0x3f, sizeof(dp));

	ios::sync_with_stdio(false);

	cin.tie(0);

	cout.tie(0);

	cin >> n >> W;

	for (int i = 1; i <= n; i++)

		cin >> w[i] >> v[i];

	ll ans = 0ll;

	dp[0] = 0;

	for (int i = 1; i <= n; i++){

		for (int j = 1e5; j >= v[i]; --j){

			dp[j] = min(dp[j], dp[j - v[i]] + w[i]);

			if (dp[j] <= W) ans = max(ans, (ll)(j));

		}

	}

	cout << ans;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 6.260531457399618x):
#include <bits/stdc++.h>

using namespace std;



int main() {

    long long n,w;    cin>>n>>w;

    long long dp[100005];

    memset(dp, 1e9+1, sizeof(dp));

    long long mx_till_now=0;

    dp[0]=0;

    for(long long i=0;i<n;i++){

        long long wi,vi;cin>>wi>>vi;

        for(long long j=mx_till_now;j>=0;j--)if( dp[j]!=1e9+1 && (dp[j] + wi==1e9+1 ||(dp[j] + wi)<=w )){

                dp[j+vi] = min ( dp[j+vi] , dp[j] + wi );

                mx_till_now=max(mx_till_now,j+vi);

        }

    }   

    cout<<mx_till_now;

    

    return 0;

}
// Slower C++ solution:
/*

    Mbak Rose Dong

*/

#include<bits/stdc++.h>

using namespace std;

long long int weight[105], value[105];

long long int dp[100100];

int main()

{

    int n, i, j, k, curr, m, ans;

    int tc;

    int w;

    scanf("%d %d",&n,&w);

    for(i=1;i<=n;i++)

    {

        scanf("%lld %lld",&weight[i], &value[i]);

    }

    for(i=1;i<=100000;i++)

        dp[i] = 1e15;

    dp[0] = 0;

    for(i=1;i<=n;i++)

    {

        for(j=100000-value[i];j>=0;j--)

        {

            dp[j+value[i]] = min(dp[j+value[i]], dp[j] + weight[i]);

        }

    }

    for(i=100000;i>=0;i--)

    {

        if(dp[i] <= w)

        {

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

            return 0;

        }

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4072176535502425x):
#include <bits/stdc++.h>

using namespace std;



int main() {

    long long n,w;    cin>>n>>w;

    long long dp[100005];

    memset(dp, 1e9+1, sizeof(dp));

    long long mx_till_now=0;

    dp[0]=0;

    for(long long i=0;i<n;i++){

        long long wi,vi;cin>>wi>>vi;

        for(long long j=mx_till_now;j>=0;j--)if( dp[j]!=1e9+1 && (dp[j] + wi==1e9+1 ||(dp[j] + wi)<=w )){

                dp[j+vi] = min ( dp[j+vi] , dp[j] + wi );

                mx_till_now=max(mx_till_now,j+vi);

        }

    }   

    cout<<mx_till_now;

    

    return 0;

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

#include<map>

#include<cmath>

#include<ctime>

#include<queue>

#include<stack>

#include<cstdio>

#include<string>

#include<vector>

#include<random>

#include<cstdlib>

#include<cstring>

#include<iomanip>

#include<iostream>

#include<algorithm>

#include<unordered_map>

#define endl "\n"

#define fi first

#define se second

#define db double

#define gcd __gcd

#define pb push_back

#define mp make_pair

#define lowbit(x) (x & (-x))

#define PII  pair<int, int> 

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

#define debug(x) cout << #x << " = " << x << endl

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

#define Rep(i, a, b) for(__typeof(a) i = a; i >= (b); i--)

#define FAST ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

template<class T> inline T qmin(T a, T b) { return a < b ? a : b; }

template<class T> inline T qmax(T a, T b) { return a > b ? a : b; }

typedef long long ll;

typedef unsigned long long ull;

const db eps = 1e-9;

const db PI = acos(-1);

const int inf = 0x3f3f3f3f;

const int mod = (int)1e9 + 7;

const int maxn = (int)1e5 + 5;//remember to modify it, No RE or MLE

const ll INF = 0x3f3f3f3f3f3f3f3f;

using namespace std;

ll qpow(ll x, ll y) { ll ret = 1; for( ; y; y >>= 1, x = (x * x) % mod) if(y & 1) ret = (ret * x) % mod; return ret; }



inline int read(){

	char ch = getchar();

	int x = 0, f = 1;

	while(ch < '0' || ch > '9'){

		if(ch == '-') f = -1;

		ch = getchar();

	}

	while('0' <= ch && ch <= '9'){

		x = x * 10 + ch - '0';

		ch = getchar();

	}

	return x * f;

}



int n, W;

int w[105], v[105];

ll dp[maxn];



int main()

{

	n = read(), W = read();

	rep(i, 1, n) w[i] = read(), v[i] = read();

	memset(dp, 0x3f, sizeof dp);

	dp[0] = 0;

	rep(i, 1, n){

		Rep(j, maxn - 1, v[i]){

			dp[j] = qmin(dp[j], dp[j-v[i]] + w[i]);

		}

	}

	Rep(i, maxn - 1, 0){

		if(dp[i] <= W) return 0 * printf("%d\n", i);

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4047928121444169x):
#include <bits/stdc++.h>

using namespace std;



int main() {

    long long n,w;    cin>>n>>w;

    long long dp[100005];

    memset(dp, 1e9+1, sizeof(dp));

    long long mx_till_now=0;

    dp[0]=0;

    for(long long i=0;i<n;i++){

        long long wi,vi;cin>>wi>>vi;

        for(long long j=mx_till_now;j>=0;j--)if( dp[j]!=1e9+1 && (dp[j] + wi==1e9+1 ||(dp[j] + wi)<=w )){

                dp[j+vi] = min ( dp[j+vi] , dp[j] + wi );

                mx_till_now=max(mx_till_now,j+vi);

        }

    }   

    cout<<mx_till_now;

    

    return 0;

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

#include <cstring>

#include <algorithm>

#include <iostream>





using namespace std;



const int MaxN = 1e5 + 5;



long long dp[MaxN];

long long w[105] , W , n;

long long v[MaxN];



int main(){

	memset(dp , 0x3f , sizeof dp);

	cin >> n >> W;

	for(int i = 1; i <= n ; i++){

		cin >> w[i] >> v[i];

	}

	dp[0] = 0;

	for(int i = 1; i <= n ; i++){

		for(int j = MaxN - 1 ; j >= v[i] ; j--){

			dp[j] = min(dp[j],dp[j - v[i]] + w[i]);

		}

	}

	for(int i = MaxN - 1 ; i >= 0 ; i--){

		if(dp[i] <= W){

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

			break;

		}

	}

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.4113861463873818x):
#include <bits/stdc++.h>

using namespace std;



int main() {

    long long n,w;    cin>>n>>w;

    long long dp[100005];

    memset(dp, 1e9+1, sizeof(dp));

    long long mx_till_now=0;

    dp[0]=0;

    for(long long i=0;i<n;i++){

        long long wi,vi;cin>>wi>>vi;

        for(long long j=mx_till_now;j>=0;j--)if( dp[j]!=1e9+1 && (dp[j] + wi==1e9+1 ||(dp[j] + wi)<=w )){

                dp[j+vi] = min ( dp[j+vi] , dp[j] + wi );

                mx_till_now=max(mx_till_now,j+vi);

        }

    }   

    cout<<mx_till_now;

    

    return 0;

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



#define pb push_back

#define se second

#define fi first



using namespace std;



int n, c[(int)1e5 + 7], a[(int)1e5 + 7], w;

long long d[(int)2e5 + 7], mx;



int main(){

    ios_base::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

    cin >> n >> w;

    for(int i = 1; i <= n; i++)

        cin >> a[i] >> c[i];

    for(int i = 1e5 + 6; i > 0; i--)

        d[i] = 1e9 + 7;

    for(int i = 1; i <= n; i++)

    for(int j = 1e5 + 6; j >= 0; j--)

        if(d[j] != 1e9 + 7)

            d[j + c[i]] = min(d[j + c[i]], d[j] + a[i]);

    for(int j = 0; j <= 1e5 + 6; j++)

        if(d[j] <= w)

            mx = j;

    cout << mx;

}


// Faster C++ solution for the same problem (measured speedup: 1.4839389768893496x):
#include <bits/stdc++.h>

using namespace std;



int main() {

    long long n,w;    cin>>n>>w;

    long long dp[100005];

    memset(dp, 1e9+1, sizeof(dp));

    long long mx_till_now=0;

    dp[0]=0;

    for(long long i=0;i<n;i++){

        long long wi,vi;cin>>wi>>vi;

        for(long long j=mx_till_now;j>=0;j--)if( dp[j]!=1e9+1 && (dp[j] + wi==1e9+1 ||(dp[j] + wi)<=w )){

                dp[j+vi] = min ( dp[j+vi] , dp[j] + wi );

                mx_till_now=max(mx_till_now,j+vi);

        }

    }   

    cout<<mx_till_now;

    

    return 0;

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

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f3f3f3f

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define pb push_back

#define ms(a, x) memset(a, x, sizeof(a))

#define forn(i, n) for(int i = 0; i < n; ++i)

#define for1(i, n) for(int i = 1; i <= n; ++i) 

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

using namespace std;

const int mod = 1e9;

const int N = 1e5 + 10;

/*=======================================================================*/



ll n, W, w[N], v[N], f[N];

ll ans = 0;

int main() {

  cin >> n >> W;

  ms(f, inf);

  f[0] = 0;

  for (int i = 1; i <= n; i++) cin >> w[i] >> v[i];



  for (int i = 1; i <= n; i++) {

     for (int l = 1e5; l >= v[i]; l--) {

        if (f[l - v[i]] + w[i] < f[l]) {

         //   debug1

            f[l] = f[l - v[i]] + w[i];

         }

         if(f[l] <= W) ans = max(ans, (ll)l);

     }

  }

  cout << ans << endl;

  return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3588269516578004x):
#include <bits/stdc++.h>

using namespace std;



int main() {

    long long n,w;    cin>>n>>w;

    long long dp[100005];

    memset(dp, 1e9+1, sizeof(dp));

    long long mx_till_now=0;

    dp[0]=0;

    for(long long i=0;i<n;i++){

        long long wi,vi;cin>>wi>>vi;

        for(long long j=mx_till_now;j>=0;j--)if( dp[j]!=1e9+1 && (dp[j] + wi==1e9+1 ||(dp[j] + wi)<=w )){

                dp[j+vi] = min ( dp[j+vi] , dp[j] + wi );

                mx_till_now=max(mx_till_now,j+vi);

        }

    }   

    cout<<mx_till_now;

    

    return 0;

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



using namespace std;

typedef long long ll;

typedef unsigned long long ull;



const int mod  = 100003;

const int maxn = 1e5+10;



ll val[105],v[105],dp[100025];

int main()

{

	int n,V;

	ll ans = 0;

	scanf("%d%d",&n,&V);

    for(int i = 1;i<=n;++i)

		scanf("%lld%lld",&v[i],&val[i]);

    memset(dp,0x3f,sizeof(dp));

    dp[0]=0;

    for(int i = 1;i<=n;++i)

    {

        for(int j = 100000;j>=val[i];j--)

            dp[j] = min(dp[j],dp[j-val[i]]+v[i]);

    }

    for(int i = 1;i<=1e5;++i)

    	if(dp[i]<=V)

    		ans = max(ans,1ll*i);

    printf("%lld\n",ans);

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.4430532970049623x):
#include <bits/stdc++.h>

using namespace std;



int main() {

    long long n,w;    cin>>n>>w;

    long long dp[100005];

    memset(dp, 1e9+1, sizeof(dp));

    long long mx_till_now=0;

    dp[0]=0;

    for(long long i=0;i<n;i++){

        long long wi,vi;cin>>wi>>vi;

        for(long long j=mx_till_now;j>=0;j--)if( dp[j]!=1e9+1 && (dp[j] + wi==1e9+1 ||(dp[j] + wi)<=w )){

                dp[j+vi] = min ( dp[j+vi] , dp[j] + wi );

                mx_till_now=max(mx_till_now,j+vi);

        }

    }   

    cout<<mx_till_now;

    

    return 0;

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

#include<cmath>

#include<cstring>

#include<cstdio>

using namespace std;

typedef long long ll;

ll a[110],b[110];

ll dp[100005];

const int maxn=100005;

int main()

{

	ll n,w;

	cin>>n>>w;

	for(ll i=0;i<n;i++){

		cin>>a[i]>>b[i];

	}

    

    memset(dp,0x3f3f3f,sizeof(dp));

    dp[0]=0;

    ll ans=0;

	for(ll i=0;i<n;i++){

		for(ll j=maxn;j>=b[i];j--){

			dp[j]=min(dp[j],dp[j-b[i]]+a[i]);

		}

	}

	

	for(ll i=0;i<maxn;i++){

		if(w>=dp[i])

			ans=max(ans,i);

	} 

	cout<<ans<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.4451889494689474x):
#include <bits/stdc++.h>

using namespace std;



int main() {

    long long n,w;    cin>>n>>w;

    long long dp[100005];

    memset(dp, 1e9+1, sizeof(dp));

    long long mx_till_now=0;

    dp[0]=0;

    for(long long i=0;i<n;i++){

        long long wi,vi;cin>>wi>>vi;

        for(long long j=mx_till_now;j>=0;j--)if( dp[j]!=1e9+1 && (dp[j] + wi==1e9+1 ||(dp[j] + wi)<=w )){

                dp[j+vi] = min ( dp[j+vi] , dp[j] + wi );

                mx_till_now=max(mx_till_now,j+vi);

        }

    }   

    cout<<mx_till_now;

    

    return 0;

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

#include<cstring> 

using namespace std;

int main()

{

	long long n,w;

	cin>>n>>w;

	long long wt[n+1],v[n+1];

	for(long long i=1;i<=n;i++)

	cin>>wt[i]>>v[i];

	long long dp[100005];

	memset(dp,0x3f,sizeof(dp));

		dp[0]=0;

	for(int i=1;i<=n;i++)

	 for(int j=100000;j>=v[i];j--)

	 dp[j]=min(dp[j],dp[j-v[i]]+wt[i]);

	long long ans=0;

	for(long long i=1;i<=100000;i++)

		if(dp[i]<=w)

		ans=max(ans,i);

		cout<<ans;

	

 } 

// Faster C++ solution for the same problem (measured speedup: 1.4442406776879113x):
#include <bits/stdc++.h>

using namespace std;



int main() {

    long long n,w;    cin>>n>>w;

    long long dp[100005];

    memset(dp, 1e9+1, sizeof(dp));

    long long mx_till_now=0;

    dp[0]=0;

    for(long long i=0;i<n;i++){

        long long wi,vi;cin>>wi>>vi;

        for(long long j=mx_till_now;j>=0;j--)if( dp[j]!=1e9+1 && (dp[j] + wi==1e9+1 ||(dp[j] + wi)<=w )){

                dp[j+vi] = min ( dp[j+vi] , dp[j] + wi );

                mx_till_now=max(mx_till_now,j+vi);

        }

    }   

    cout<<mx_till_now;

    

    return 0;

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

#include<cstring>

#include<cstdio>

using namespace std;

typedef long long ll;

const int maxn = 1000010;

const int inf = 0x3f3f3f3f;

ll dp[maxn];

ll n, W;

ll v[maxn];

ll w[maxn];

inline ll max(ll x, ll y) {

	return x > y ? x : y;

}

inline ll min(ll x, ll y) {

	return x < y ? x : y;

}

int main() {

	memset(dp, 0x3f, sizeof(dp));

	ios::sync_with_stdio(false);

	cin.tie(0);

	cout.tie(0);

	cin >> n >> W;

	for (int i = 1; i <= n; i++)

		cin >> w[i] >> v[i];

	ll ans = 0ll;

	dp[0] = 0;

	for (int i = 1; i <= n; i++){

		for (int j = 1e5; j >= v[i]; --j){

			dp[j] = min(dp[j], dp[j - v[i]] + w[i]);

			if (dp[j] <= W) ans = max(ans, (ll)(j));

		}

	}

	cout << ans;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.448534399513893x):
#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ull;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

const int maxn = 1e5 + 9;

inline int read()

{

    int data = 0, f = 1;

    char ch = getchar();

    while (!isdigit(ch))

    {

        if (ch == '-')

            f = -1;

        ch = getchar();

    }

    while (isdigit(ch))

    {

        data = (data << 3) + (data << 1) + ch - '0';

        ch = getchar();

    }

    return f * data;

}



int v[109], p[1009];

ull dp[maxn];

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("data.txt","w",stdout);

    //std::ios::sync_with_stdio(false);

    //std::cin.tie(0);

    int n = read(), w = read();

    for (int i = 1; i <= n; ++i)

        p[i] = read(), v[i] = read();

    for (ull i = 1; i < maxn; ++i)

        dp[i] = INF;

    for (int i = 1; i <= n; ++i)

        for (int j = maxn - 1; j >= v[i]; --j)

            dp[j] = min(dp[j], dp[j - v[i]] + p[i]);

    for (int i = maxn - 1; i >= 0; --i)

        if (dp[i] <= w)

        {

            cout << i << '\n';

            break;

        }

    //fclose(stdin);

    //fclose(stdout);

    return 0;

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

#include<cstring>

#include<cstdio>

using namespace std;

typedef long long ll;

const int maxn = 1000010;

const int inf = 0x3f3f3f3f;

ll dp[maxn];

ll n, W;

ll v[maxn];

ll w[maxn];

inline ll max(ll x, ll y) {

	return x > y ? x : y;

}

inline ll min(ll x, ll y) {

	return x < y ? x : y;

}

int main() {

	memset(dp, 0x3f, sizeof(dp));

	ios::sync_with_stdio(false);

	cin.tie(0);

	cout.tie(0);

	cin >> n >> W;

	for (int i = 1; i <= n; i++)

		cin >> w[i] >> v[i];

	ll ans = 0ll;

	dp[0] = 0;

	for (int i = 1; i <= n; i++){

		for (int j = 1e5; j >= v[i]; --j){

			dp[j] = min(dp[j], dp[j - v[i]] + w[i]);

			if (dp[j] <= W) ans = max(ans, (ll)(j));

		}

	}

	cout << ans;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.760352106371866x):
#include <iostream>

#include <cstring>

using namespace std;



typedef long long ll;



const int maxn = 105, maxw = 1e9 + 5, maxv = 1e3 + 5;



int main(int argc, const char * argv[])

{

    ll n, W, w[maxn], v[maxn], dp[maxv*maxn], ans = 0;

    //memset(dp, maxw, sizeof dp);

    for (int i = 0; i < maxv * maxn; i++)

        dp[i] = maxw;

    cin >> n >> W;

    for (int i = 0; i < n; i++)

        scanf("%lld %lld", &w[i], &v[i]);



    dp[0] = 0;

    for (int i = 0; i < n; i++) {

        for (int j = maxv * maxn - 1; j >= v[i]; j--) {

            dp[j] = min(dp[j], dp[j-v[i]] + w[i]);

            if (dp[j] <= W)

                ans = max(ans, (ll)j);

        }

    }

    

    cout << ans << endl;

    return 0;

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

#include <vector>

#include <algorithm>

#include <numeric>

#include <cmath>

#include <set>

#include <tuple>

#include <functional>

#include <map>

#include <unordered_map>

#include <queue>

#include <tuple>



using namespace std;



#define endl '\n'



using ull = unsigned long int;

using ll = long long;



int main() {

  ios_base::sync_with_stdio(false);

  cin.tie(nullptr);

  const int precision = 13;

  cout.precision(precision);

  cout.setf(ios_base::fixed);



  int n, W; cin >> n >> W;

  vector<ll> dp(1e5+10, W+1);



  dp[0] = 0;

  for (int i = 0; i < n; ++i) {

    ll w, v; cin >>  w >> v;

    for (int j = 1e5; j - v >= 0; --j) {

      dp[j] = min(dp[j], dp[j-v] + w);

    }

  }



  int ans = 0;

  for (int i = 1; i <= 1e5; ++i) {

    if (dp[i] <= W) {

      ans = max(ans, i);

    }

  }

  cout << ans << endl;



  return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.4688114399139747x):
#include <bits/stdc++.h>



using namespace std;



int main() {

    long long N, W, V = 0;

    cin >> N >> W;



    vector<int> w(N), v(N);

    for (int i = 0; i < N; i++) {

        cin >> w[i] >> v[i];

        V += v[i];

    }



    vector<long long> ks(V + 1, INT_MAX);

    ks[0] = 0;



    for (int i = 0; i < N; i++)

        for (int j = V; j >= v[i]; j--) ks[j] = min(ks[j - v[i]] + w[i], ks[j]);



    for (int i = V; i >= 0; i--) {

        if (ks[i] <= W) {

            cout << i;

            break;

        }

    }



    return 0;

}
// Slower C++ solution:
/**

 *  Author : Coda

 *  Title : OSU!

 *  Date : 2019-08-20

 */

#include <bits/stdc++.h>

using namespace std ;



typedef long long ll ;

typedef unsigned long long ull ;

typedef double DB ;

const long long mod = 1e9+7 ;

const double pi = acos(-1) ;

#define PB push_back

#define MP make_pair

#define lb lower_bound

#define ub upper_bound

#define fs first

#define sd second

#define pii pair<int,int>

#define INF 0x3f3f3f3f

#define LLINF 0x3f3f3f3f3f3f3f3f

#define debug(x) cout << #x << " = " << x << endl

#define debug2(x,y) cout << #x << " = " << x << ' ' << #y << " = " << y << endl

#define debug3(x,y,z) cout << #x << " = " << x << ' ' << #y << " = " << y << ' ' << #z << " = " << z << endl

//#pragma comment(linker,"/STACK:10240000,10240000")

//#include <random>

//mt19937 rand_(time(0));

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

ll qpow(ll a,ll b,ll c){ll t=1;while(b){if(b&1)t=t*a%c;a=a*a%c;b>>=1;}return t;}

ll inv(ll x){return qpow(x,mod-2,mod);}

int test = 1 ;



const int maxn = 5 + 3e5 ;

const int maxm = 5 + 1e6 ;

// Write the variables here



int N,W ;

ll v[maxn],w[maxn] ;

ll dp[maxn] ;



//#define TEST

int main(){

#ifdef TEST

    freopen("in.txt","r",stdin) ;

#endif

    while(test--){

        // code begins

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



        ll sum = 0 ;

        for(int i=1 ; i<=N ; i++){

            scanf("%lld%lld",&w[i],&v[i]) ;

        }

        for(int i=1 ; i<=100000 ; i++){

            dp[i] = LLINF ;

        }

        dp[0] = 0;

        ll ans = 0 ;

        for(int i=1 ; i<=N ; i++){

            for(int j=100000 ; j>=v[i] ; j--){

                dp[j] = min(dp[j],dp[j-v[i]]+w[i]);

            }

        }

        for(int i=1 ; i<=100000 ; i++){

            if(dp[i]<=W) ans = max(ans,i*1ll) ;

        }

        printf("%lld\n",ans) ;



    }

#ifdef TEST

    cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;

#endif

    return 0 ;

}


// Faster C++ solution for the same problem (measured speedup: 2.5607274976014374x):
#include <bits/stdc++.h>



using namespace std;



int main() {

    long long N, W, V = 0;

    cin >> N >> W;



    vector<int> w(N), v(N);

    for (int i = 0; i < N; i++) {

        cin >> w[i] >> v[i];

        V += v[i];

    }



    vector<long long> ks(V + 1, INT_MAX);

    ks[0] = 0;



    for (int i = 0; i < N; i++)

        for (int j = V; j >= v[i]; j--) ks[j] = min(ks[j - v[i]] + w[i], ks[j]);



    for (int i = V; i >= 0; i--) {

        if (ks[i] <= W) {

            cout << i;

            break;

        }

    }



    return 0;

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

#include<cstring>

#include<cstdio>

using namespace std;

typedef long long ll;

const int maxn = 1000010;

const int inf = 0x3f3f3f3f;

ll dp[maxn];

ll n, W;

ll v[maxn];

ll w[maxn];

inline ll max(ll x, ll y) {

	return x > y ? x : y;

}

inline ll min(ll x, ll y) {

	return x < y ? x : y;

}

int main() {

	memset(dp, 0x3f, sizeof(dp));

	ios::sync_with_stdio(false);

	cin.tie(0);

	cout.tie(0);

	cin >> n >> W;

	for (int i = 1; i <= n; i++)

		cin >> w[i] >> v[i];

	ll ans = 0ll;

	dp[0] = 0;

	for (int i = 1; i <= n; i++){

		for (int j = 1e5; j >= v[i]; --j){

			dp[j] = min(dp[j], dp[j - v[i]] + w[i]);

			if (dp[j] <= W) ans = max(ans, (ll)(j));

		}

	}

	cout << ans;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 11.142874108659138x):
#include <bits/stdc++.h>



using namespace std;



int main() {

    long long N, W, V = 0;

    cin >> N >> W;



    vector<int> w(N), v(N);

    for (int i = 0; i < N; i++) {

        cin >> w[i] >> v[i];

        V += v[i];

    }



    vector<long long> ks(V + 1, INT_MAX);

    ks[0] = 0;



    for (int i = 0; i < N; i++)

        for (int j = V; j >= v[i]; j--) ks[j] = min(ks[j - v[i]] + w[i], ks[j]);



    for (int i = V; i >= 0; i--) {

        if (ks[i] <= W) {

            cout << i;

            break;

        }

    }



    return 0;

}
// Slower C++ solution:
/*

    Mbak Rose Dong

*/

#include<bits/stdc++.h>

using namespace std;

long long int weight[105], value[105];

long long int dp[100100];

int main()

{

    int n, i, j, k, curr, m, ans;

    int tc;

    int w;

    scanf("%d %d",&n,&w);

    for(i=1;i<=n;i++)

    {

        scanf("%lld %lld",&weight[i], &value[i]);

    }

    for(i=1;i<=100000;i++)

        dp[i] = 1e15;

    dp[0] = 0;

    for(i=1;i<=n;i++)

    {

        for(j=100000-value[i];j>=0;j--)

        {

            dp[j+value[i]] = min(dp[j+value[i]], dp[j] + weight[i]);

        }

    }

    for(i=100000;i>=0;i--)

    {

        if(dp[i] <= w)

        {

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

            return 0;

        }

    }

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.5046514443210084x):
#include <bits/stdc++.h>



using namespace std;



int main() {

    long long N, W, V = 0;

    cin >> N >> W;



    vector<int> w(N), v(N);

    for (int i = 0; i < N; i++) {

        cin >> w[i] >> v[i];

        V += v[i];

    }



    vector<long long> ks(V + 1, INT_MAX);

    ks[0] = 0;



    for (int i = 0; i < N; i++)

        for (int j = V; j >= v[i]; j--) ks[j] = min(ks[j - v[i]] + w[i], ks[j]);



    for (int i = V; i >= 0; i--) {

        if (ks[i] <= W) {

            cout << i;

            break;

        }

    }



    return 0;

}