stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include <bits/stdc++.h>

#define ll long long

#define ull unsigned ll

#define ld long double

#define pll pair<ll, ll>

#define pii pair<int, int>

#define f first

#define se second

#define pb push_back

#define ft fflush(stdout)

 

 

using namespace std;

 

 

const int N = 1e5 + 123;

const int PLS = 3e5 + 123;

const int mod = 1e9 + 7;

const ll inf = 1e16;





ll n, a[N], b[N], w, dp[110][N];





int main() {

    ios_base::sync_with_stdio(NULL);

    cin.tie(NULL); cout.tie(NULL);

    cin >> n >> w;

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

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

    dp[0][0] = 0;

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

        dp[0][i] = inf;

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

        for (int j = 0; j <= 100000; j++) {

            dp[i][j] = dp[i - 1][j];

            if (j >= b[i])

                dp[i][j] = min(dp[i][j], dp[i - 1][j - b[i]] + a[i]);

        }

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

        if (dp[n][i] <= w)

            return cout << i, 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.217996589991311x):
//I

//01背包

#include<cstdio>

#include<iostream> 

#include<cstring>

using namespace std;

int w[110],c[110];

long long f[100010];//f[v]表示不超过v价值的最小体积 

int main()

{

	for(int i=0;i<100010;i++)

		f[i]=1000000010;

	long long m,n,flag=0; 

	scanf("%d%d",&n,&m);//背包容量m和物品数量n 

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

		scanf("%d%d",&w[i],&c[i]);//重量和价值 

	f[0]=0;

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

		for(int v=100010;v>=c[i];v--)

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

			{

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

				if(f[v]<=m&&v>flag)

					flag=v;

			}

	printf("%lld",flag);

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

#include <iomanip>

#include <cmath>

#include <cstdio>

#include <cstring>

#include <ctime>

#include <string>

#include <stdio.h>

#include <set>

#include <map>

#include <algorithm>

#include <queue>

#include <vector>

#define ll long long

#define pb push_back

#define mp make_pair

#define f first

#define s second



using namespace std;



const int inf = 1e9 + 7;

const long long INF = 1e18 + 7;

const int maxn = 2e5 + 7;



ll n, s, w[maxn], a[maxn];

ll dp[maxn], mx;



int main(){

	ios_base::sync_with_stdio(NULL);

	cin.tie(0);

	cout.tie(0);

	cin >> n >> s;

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

		dp[i] = inf;

	dp[0] = 0;

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

		cin >> w[i] >> a[i];

		for (ll j = 100000; j >= 0; j--){

			if (dp[j] != inf && dp[j] + w[i] <= s){

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

			}

		}

	}

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

		if (dp[i] != inf)

			mx = max(mx, i);

	cout << mx;

	return 0;

} 

// Faster C++ solution for the same problem (measured speedup: 1.4683523109612462x):
//I

//01背包

#include<cstdio>

#include<iostream> 

#include<cstring>

using namespace std;

int w[110],c[110];

long long f[100010];//f[v]表示不超过v价值的最小体积 

int main()

{

	for(int i=0;i<100010;i++)

		f[i]=1000000010;

	long long m,n,flag=0; 

	scanf("%d%d",&n,&m);//背包容量m和物品数量n 

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

		scanf("%d%d",&w[i],&c[i]);//重量和价值 

	f[0]=0;

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

		for(int v=100010;v>=c[i];v--)

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

			{

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

				if(f[v]<=m&&v>flag)

					flag=v;

			}

	printf("%lld",flag);

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

using namespace std;

const int MX = 200005;

long long dp[MX];  // dp[i]表示价值为i时的最小体积

int main()

{

	int n,w;

	cin >> n >> w;

	for(int i = 0; i < MX; ++i) dp[i] = 1001002003004005006ll;

	dp[0] = 0;

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

	{

		int a,b;

		cin >> a >> b;

		for (int j = MX-1; j >= b; --j) 

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

	}

	int ans = 0;

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

		if (dp[i] <= w) ans = i;

	cout << ans << endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.9383964529666615x):
//I

//01背包

#include<cstdio>

#include<iostream> 

#include<cstring>

using namespace std;

int w[110],c[110];

long long f[100010];//f[v]表示不超过v价值的最小体积 

int main()

{

	for(int i=0;i<100010;i++)

		f[i]=1000000010;

	long long m,n,flag=0; 

	scanf("%d%d",&n,&m);//背包容量m和物品数量n 

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

		scanf("%d%d",&w[i],&c[i]);//重量和价值 

	f[0]=0;

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

		for(int v=100010;v>=c[i];v--)

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

			{

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

				if(f[v]<=m&&v>flag)

					flag=v;

			}

	printf("%lld",flag);

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

using namespace std;

#define ll long long

#define ull unsigned long long

#define mst(a, b) memset((a), (b), sizeof(a))

#define mp(a, b) make_pair(a, b)

#define pi acos(-1)

#define pii pair<int, int>

#define pb push_back

const int INF = 0x3f3f3f3f;

const double eps = 1e-10;

const int maxn = 1e5 + 7;

const int maxm = 5e5 + 10;

ll w[maxn], v[maxn];

ll dp[maxn];

ll ans = 0;

int main()

{

    ios::sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

    int n, m;

    cin >> n >> m;

    mst(dp, INF);

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

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

    dp[0] = 0;

    for (int i = 0; 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] <= m)

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

        }

    }

    cout << ans << endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.2434909938798517x):
//I

//01背包

#include<cstdio>

#include<iostream> 

#include<cstring>

using namespace std;

int w[110],c[110];

long long f[100010];//f[v]表示不超过v价值的最小体积 

int main()

{

	for(int i=0;i<100010;i++)

		f[i]=1000000010;

	long long m,n,flag=0; 

	scanf("%d%d",&n,&m);//背包容量m和物品数量n 

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

		scanf("%d%d",&w[i],&c[i]);//重量和价值 

	f[0]=0;

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

		for(int v=100010;v>=c[i];v--)

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

			{

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

				if(f[v]<=m&&v>flag)

					flag=v;

			}

	printf("%lld",flag);

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

#include <algorithm>

#include <cmath>

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <iomanip>

#include <string>

#include <set>

#include <vector>

#include <queue>

#include <stack>

using namespace std;

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

#define fore(i, a, n) for (long long i = a; i < n; i++)

#define ford(i, a, n) for (long long i = n - 1; i >= a; i--)

#define si(a) scanf("%d", &a)

#define sii(a, b) scanf("%d%d", &a, &b)

#define debug(a) cout << a << endl

#define endl '\n'

#define pi acos(-1.0)

#define tr t[root]

#define lson t[root << 1]

#define rson t[root << 1 | 1]

#define IO ios::sync_with_stdio(false), cin.tie(0)

const double eps = 1e-8;

inline int sgn(const double &x) { return x < -eps ? -1 : x > eps; }

typedef long long ll;

const int inf = 0x3f3f3f3f;

const int maxn=1000100;

#define maxn0 1e9+7;

//=================================================



ll sum;

ll w[maxn];

ll dp[maxn];

ll v[maxn];



// void init() {

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

//         w[i]=0;

//         v[i]=0;

//     }

// }

int main() {

    IO;

    ll n,W;

    cin>>n>>W;

    // sum=0;

    // init();

    memset(dp,inf,sizeof(dp));

    dp[0]=0;

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

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

        // sum+=v[i];

    }

    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]);

        }

    }

    ll ans=-1;

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

        if(dp[i]<=W)

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

    }

    cout<<ans<<endl;



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 4.541607857136762x):
//I

//01背包

#include<cstdio>

#include<iostream> 

#include<cstring>

using namespace std;

int w[110],c[110];

long long f[100010];//f[v]表示不超过v价值的最小体积 

int main()

{

	for(int i=0;i<100010;i++)

		f[i]=1000000010;

	long long m,n,flag=0; 

	scanf("%d%d",&n,&m);//背包容量m和物品数量n 

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

		scanf("%d%d",&w[i],&c[i]);//重量和价值 

	f[0]=0;

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

		for(int v=100010;v>=c[i];v--)

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

			{

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

				if(f[v]<=m&&v>flag)

					flag=v;

			}

	printf("%lld",flag);

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

#define ll long long

#define ull unsigned ll

#define ld long double

#define pll pair<ll, ll>

#define pii pair<int, int>

#define f first

#define se second

#define pb push_back

#define ft fflush(stdout)

 

 

using namespace std;

 

 

const int N = 1e5 + 123;

const int PLS = 3e5 + 123;

const int mod = 1e9 + 7;

const ll inf = 1e16;





ll n, a[N], b[N], w, dp[110][N];





int main() {

    ios_base::sync_with_stdio(NULL);

    cin.tie(NULL); cout.tie(NULL);

    cin >> n >> w;

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

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

    dp[0][0] = 0;

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

        dp[0][i] = inf;

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

        for (int j = 0; j <= 100000; j++) {

            dp[i][j] = dp[i - 1][j];

            if (j >= b[i])

                dp[i][j] = min(dp[i][j], dp[i - 1][j - b[i]] + a[i]);

        }

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

        if (dp[n][i] <= w)

            return cout << i, 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8734292766249683x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long

ll dp[110000];

ll ww[110],v[110];

int main()

{

    int n,w;cin>>n>>w;

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

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

    dp[0]=0;

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

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

            if(dp[j-v[i]]+ww[i]<=w)

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

    int maxx=0;

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

        if(dp[i]<=w)maxx=i;

    cout<<maxx<<endl;

    return 0;

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

#include <iomanip>

#include <cmath>

#include <cstdio>

#include <cstring>

#include <ctime>

#include <string>

#include <stdio.h>

#include <set>

#include <map>

#include <algorithm>

#include <queue>

#include <vector>

#define ll long long

#define pb push_back

#define mp make_pair

#define f first

#define s second



using namespace std;



const int inf = 1e9 + 7;

const long long INF = 1e18 + 7;

const int maxn = 2e5 + 7;



ll n, s, w[maxn], a[maxn];

ll dp[maxn], mx;



int main(){

	ios_base::sync_with_stdio(NULL);

	cin.tie(0);

	cout.tie(0);

	cin >> n >> s;

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

		dp[i] = inf;

	dp[0] = 0;

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

		cin >> w[i] >> a[i];

		for (ll j = 100000; j >= 0; j--){

			if (dp[j] != inf && dp[j] + w[i] <= s){

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

			}

		}

	}

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

		if (dp[i] != inf)

			mx = max(mx, i);

	cout << mx;

	return 0;

} 

// Faster C++ solution for the same problem (measured speedup: 1.2402427578869741x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long

ll dp[110000];

ll ww[110],v[110];

int main()

{

    int n,w;cin>>n>>w;

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

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

    dp[0]=0;

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

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

            if(dp[j-v[i]]+ww[i]<=w)

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

    int maxx=0;

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

        if(dp[i]<=w)maxx=i;

    cout<<maxx<<endl;

    return 0;

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

using namespace std;

const int MX = 200005;

long long dp[MX];  // dp[i]表示价值为i时的最小体积

int main()

{

	int n,w;

	cin >> n >> w;

	for(int i = 0; i < MX; ++i) dp[i] = 1001002003004005006ll;

	dp[0] = 0;

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

	{

		int a,b;

		cin >> a >> b;

		for (int j = MX-1; j >= b; --j) 

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

	}

	int ans = 0;

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

		if (dp[i] <= w) ans = i;

	cout << ans << endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.6372652154113378x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long

ll dp[110000];

ll ww[110],v[110];

int main()

{

    int n,w;cin>>n>>w;

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

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

    dp[0]=0;

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

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

            if(dp[j-v[i]]+ww[i]<=w)

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

    int maxx=0;

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

        if(dp[i]<=w)maxx=i;

    cout<<maxx<<endl;

    return 0;

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

#include <algorithm>

#include <cmath>

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <iomanip>

#include <string>

#include <set>

#include <vector>

#include <queue>

#include <stack>

using namespace std;

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

#define fore(i, a, n) for (long long i = a; i < n; i++)

#define ford(i, a, n) for (long long i = n - 1; i >= a; i--)

#define si(a) scanf("%d", &a)

#define sii(a, b) scanf("%d%d", &a, &b)

#define debug(a) cout << a << endl

#define endl '\n'

#define pi acos(-1.0)

#define tr t[root]

#define lson t[root << 1]

#define rson t[root << 1 | 1]

#define IO ios::sync_with_stdio(false), cin.tie(0)

const double eps = 1e-8;

inline int sgn(const double &x) { return x < -eps ? -1 : x > eps; }

typedef long long ll;

const int inf = 0x3f3f3f3f;

const int maxn=1000100;

#define maxn0 1e9+7;

//=================================================



ll sum;

ll w[maxn];

ll dp[maxn];

ll v[maxn];



// void init() {

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

//         w[i]=0;

//         v[i]=0;

//     }

// }

int main() {

    IO;

    ll n,W;

    cin>>n>>W;

    // sum=0;

    // init();

    memset(dp,inf,sizeof(dp));

    dp[0]=0;

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

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

        // sum+=v[i];

    }

    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]);

        }

    }

    ll ans=-1;

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

        if(dp[i]<=W)

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

    }

    cout<<ans<<endl;



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 3.8360659168296234x):
#include<bits/stdc++.h>

using namespace std;

#define ll long long

ll dp[110000];

ll ww[110],v[110];

int main()

{

    int n,w;cin>>n>>w;

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

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

    dp[0]=0;

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

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

            if(dp[j-v[i]]+ww[i]<=w)

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

    int maxx=0;

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

        if(dp[i]<=w)maxx=i;

    cout<<maxx<<endl;

    return 0;

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

 

#define ll long long

#define X first

#define Y second

#define MP make_pair

#define ld long double

 

using namespace std;

 

const int N = 5e5 + 123;

const ll mod = 1e9 + 7;



ll dp[N], res, was[N];

int n, maxw;



int main (){

	ios::sync_with_stdio(0);

	cin.tie(0);

	cout.tie(0);



	cin >> n >> maxw;



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

		dp[i] = N * 1LL * N;



	was[0] = 1;

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

		ll x, y;

		cin >> x >> y;

		for(int j = 100000;j >= y;j--){

			if(!was[j - y])

				continue;

			was[j] = 1;

			dp[j] = min(dp[j], dp[j - y] + x);

		}

	}



	for(int i = 0;i <= 100000;i++){

		if(dp[i] <= maxw){

			res = i;

		}

	}



	cout << res;

}

// Faster C++ solution for the same problem (measured speedup: 2.716751061334958x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

  ios::sync_with_stdio(0);cin.tie(0);ios_base::sync_with_stdio(0);

   int n,c,sum=0,ans;

   cin>>n>>c;

   int v[n+5],w[n+5];

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

   {

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

       sum+=v[i];

   }

   long long dp[n+5][sum+10];

   for(int i=0;i<n+5;i++)

   {

       for(int j=0;j<sum+10;j++)

       {

           if(j==0)

            dp[i][j]=0;

           else

            dp[i][j]=1e18;

       }

   }

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

   {

       for(int j=1;j<=sum;j++)

       {

           if(v[i]<=j)

            dp[i][j]=min(dp[i-1][j],dp[i-1][j-v[i]]+w[i]);

           else

            dp[i][j]=dp[i-1][j];

       }

   }

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

   {

       for(int j=0;j<=sum;j++)

       {

           if(dp[i][j]<=c)

            ans=j;

       }

   }

   cout<<ans;

    return 0;

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

#define ll long long

#define ull unsigned ll

#define ld long double

#define pll pair<ll, ll>

#define pii pair<int, int>

#define f first

#define se second

#define pb push_back

#define ft fflush(stdout)

 

 

using namespace std;

 

 

const int N = 1e5 + 123;

const int PLS = 3e5 + 123;

const int mod = 1e9 + 7;

const ll inf = 1e16;





ll n, a[N], b[N], w, dp[110][N];





int main() {

    ios_base::sync_with_stdio(NULL);

    cin.tie(NULL); cout.tie(NULL);

    cin >> n >> w;

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

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

    dp[0][0] = 0;

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

        dp[0][i] = inf;

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

        for (int j = 0; j <= 100000; j++) {

            dp[i][j] = dp[i - 1][j];

            if (j >= b[i])

                dp[i][j] = min(dp[i][j], dp[i - 1][j - b[i]] + a[i]);

        }

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

        if (dp[n][i] <= w)

            return cout << i, 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.083591353730341x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

  ios::sync_with_stdio(0);cin.tie(0);ios_base::sync_with_stdio(0);

   int n,c,sum=0,ans;

   cin>>n>>c;

   int v[n+5],w[n+5];

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

   {

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

       sum+=v[i];

   }

   long long dp[n+5][sum+10];

   for(int i=0;i<n+5;i++)

   {

       for(int j=0;j<sum+10;j++)

       {

           if(j==0)

            dp[i][j]=0;

           else

            dp[i][j]=1e18;

       }

   }

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

   {

       for(int j=1;j<=sum;j++)

       {

           if(v[i]<=j)

            dp[i][j]=min(dp[i-1][j],dp[i-1][j-v[i]]+w[i]);

           else

            dp[i][j]=dp[i-1][j];

       }

   }

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

   {

       for(int j=0;j<=sum;j++)

       {

           if(dp[i][j]<=c)

            ans=j;

       }

   }

   cout<<ans;

    return 0;

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

#include <iomanip>

#include <cmath>

#include <cstdio>

#include <cstring>

#include <ctime>

#include <string>

#include <stdio.h>

#include <set>

#include <map>

#include <algorithm>

#include <queue>

#include <vector>

#define ll long long

#define pb push_back

#define mp make_pair

#define f first

#define s second



using namespace std;



const int inf = 1e9 + 7;

const long long INF = 1e18 + 7;

const int maxn = 2e5 + 7;



ll n, s, w[maxn], a[maxn];

ll dp[maxn], mx;



int main(){

	ios_base::sync_with_stdio(NULL);

	cin.tie(0);

	cout.tie(0);

	cin >> n >> s;

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

		dp[i] = inf;

	dp[0] = 0;

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

		cin >> w[i] >> a[i];

		for (ll j = 100000; j >= 0; j--){

			if (dp[j] != inf && dp[j] + w[i] <= s){

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

			}

		}

	}

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

		if (dp[i] != inf)

			mx = max(mx, i);

	cout << mx;

	return 0;

} 

// Faster C++ solution for the same problem (measured speedup: 3.3654258739242686x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

  ios::sync_with_stdio(0);cin.tie(0);ios_base::sync_with_stdio(0);

   int n,c,sum=0,ans;

   cin>>n>>c;

   int v[n+5],w[n+5];

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

   {

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

       sum+=v[i];

   }

   long long dp[n+5][sum+10];

   for(int i=0;i<n+5;i++)

   {

       for(int j=0;j<sum+10;j++)

       {

           if(j==0)

            dp[i][j]=0;

           else

            dp[i][j]=1e18;

       }

   }

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

   {

       for(int j=1;j<=sum;j++)

       {

           if(v[i]<=j)

            dp[i][j]=min(dp[i-1][j],dp[i-1][j-v[i]]+w[i]);

           else

            dp[i][j]=dp[i-1][j];

       }

   }

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

   {

       for(int j=0;j<=sum;j++)

       {

           if(dp[i][j]<=c)

            ans=j;

       }

   }

   cout<<ans;

    return 0;

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

 

#define int long long

//#pragma GCC optimize("Ofast")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")

 

 

#define file(s) freopen(s".in","r",stdin); freopen(s".out","w",stdout);

#define forev(i, b, a) for(int i = (b); i >= (a); --i)

#define forn(i, a, b) for(int i = (a); i <= (b); ++i)

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

#define sz(s) (int)s.size()

#define pb push_back

#define ppb pop_back

#define mp make_pair

#define s second

#define f first

 

 

using namespace std;

 

 

typedef pair < long long, long long > pll;    

typedef pair < int, int > pii;

typedef unsigned long long ull;         

typedef vector < pii > vpii;

typedef vector < int > vi;

typedef long double ldb; 

typedef long long ll;  

typedef double db;                         

 

 

const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1}, block = 555;

const pii base = mp(1171, 3307), Mod = mp(1e9 + 7, 1e9 + 9);

const int inf = 1e9, N = 1e5 + 48, mod = 1e9 + 7, maxn = 102;

const db eps = 1e-12, pi = 3.14159265359;

const ll INF = 1e18;





int n, m, w[maxn], c[maxn], dp[N], M = N - 12;



main () {

	cin >> n >> m;

	forn (i, 1, n)

		cin >> w[i] >> c[i];





	dp[0] = 0;

	forn (i, 1, M)

		dp[i] = INF;





	forn (i, 1, n)

		forev (j, M, c[i])

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



	forev (i, M, 0)

		if (dp[i] <= m)  {

			cout << i << endl;

			return 0;

		}

}                       

// Faster C++ solution for the same problem (measured speedup: 2.505850527855152x):
#include <bits/stdc++.h>

using namespace std;

int main()

{

  ios::sync_with_stdio(0);cin.tie(0);ios_base::sync_with_stdio(0);

   int n,c,sum=0,ans;

   cin>>n>>c;

   int v[n+5],w[n+5];

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

   {

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

       sum+=v[i];

   }

   long long dp[n+5][sum+10];

   for(int i=0;i<n+5;i++)

   {

       for(int j=0;j<sum+10;j++)

       {

           if(j==0)

            dp[i][j]=0;

           else

            dp[i][j]=1e18;

       }

   }

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

   {

       for(int j=1;j<=sum;j++)

       {

           if(v[i]<=j)

            dp[i][j]=min(dp[i-1][j],dp[i-1][j-v[i]]+w[i]);

           else

            dp[i][j]=dp[i-1][j];

       }

   }

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

   {

       for(int j=0;j<=sum;j++)

       {

           if(dp[i][j]<=c)

            ans=j;

       }

   }

   cout<<ans;

    return 0;

}