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



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

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

#define fore(i, l, r) for (int i = int(l); i <= int(r); ++i)

#define ford(i, n) for (int i = int(n)-1; i >= 0; --i)

#define pb push_back

#define eb emplace_back

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

#define F first

#define S second

#define endl '\n'

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

#define de2(x,y) cout<<#x<<" = "<<(x) <<' '<< #y<<" = "<<y<<endl;



using namespace std;



typedef long long ll;

typedef pair<int, int> pii;

const int INF = 0x3f3f3f3f;

mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());



const int N=1e5+5;

ll dp[N];

int v[105],w[105];

int main() {



    ios::sync_with_stdio(false);

    cin.tie(nullptr);

	int N,W;

	cin>>N>>W;

	int sum=0;

	forn(i,N){

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

		sum+=v[i];

	}

	ms(dp,INF);

	dp[0]=0;

	forn(i,N){

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

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

		}

	}

	ll ans=0;

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

		if(dp[i]<=W){

			ans=i;

			break;

		}

	}

	cout<<ans;

	return 0;

}




// Faster C++ solution for the same problem (measured speedup: 1.3059221002296642x):
#include<bits/stdc++.h>

using namespace::std;



const int MAX = 100000 + 5;

const int inf = 1e9 + 10;



int n;

int C;

int memo[MAX];



int main(){

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

	int sum = 0;

	for(int i = 1; i < MAX; i++) memo[i] = inf;

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

		int w, v;

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

		sum += v;

		for(int j = sum; j >= v; j--){

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

		}

	}

	int ans = 0;

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

		if(memo[i] <= C) ans = i;

	}

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

	return 0;

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

using namespace std;

const int N = 1e5+100;

const int M = 200;

typedef long long ll;

int a[N],b[N],n,m;

ll f[N];



int main(){

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

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

		scanf("%d%d",&a[i],&b[i]);

	memset(f, 0x3f, sizeof f);

	f[0] = 0;

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

		for (int j = N - 10; j >= 0; j--)

			if (j - b[i] >= 0) f[j] = min(f[j], f[j-b[i]] + a[i]);



	int ans = 0;

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

		if (f[i] <= m) ans = i;

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

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 2.204639838689791x):
#include<bits/stdc++.h>

using namespace::std;



const int MAX = 100000 + 5;

const int inf = 1e9 + 10;



int n;

int C;

int memo[MAX];



int main(){

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

	int sum = 0;

	for(int i = 1; i < MAX; i++) memo[i] = inf;

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

		int w, v;

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

		sum += v;

		for(int j = sum; j >= v; j--){

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

		}

	}

	int ans = 0;

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

		if(memo[i] <= C) ans = i;

	}

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

	return 0;

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

#include<cstdlib>

#include<cstdio>

#include<algorithm>

#include<cmath>

#include<cstring>

#include<stack>

#define LL long long

using namespace std;

LL n,m;

LL dp[100011];

LL x,y,ans;

LL read()

{

	LL _=0,__=1;char ch=getchar();

	while(ch<'0'||ch>'9') {if(ch=='-')__=-1;ch=getchar();}

	while(ch>='0'&&ch<='9'){_=_*10+ch-'0';ch=getchar();}

	return _*__;

}

int main()

{

	n=read();m=read(); 

	for(int i=1;i<=100001;i++)

	dp[i]=1e9+7;

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

	{

		x=read();y=read();

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

		{

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

		}

	}

	for(int i=1;i<=100001;i++)

	{

		if(dp[i]<=m)

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

	}

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

	return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.834212365122826x):
#include<bits/stdc++.h>

using namespace::std;



const int MAX = 100000 + 5;

const int inf = 1e9 + 10;



int n;

int C;

int memo[MAX];



int main(){

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

	int sum = 0;

	for(int i = 1; i < MAX; i++) memo[i] = inf;

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

		int w, v;

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

		sum += v;

		for(int j = sum; j >= v; j--){

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

		}

	}

	int ans = 0;

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

		if(memo[i] <= C) ans = i;

	}

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

	return 0;

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

#define MAX 100005

#define pb push_back

#define mp make_pair

#define pii pair<long long,long long>

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

#define p1(a) cout<<"Check "<<a<<endl;

#define p2(a,b) cout<<"Check "<<a<<' '<<b<<endl;

#define p3(a,b,c) cout<<"Check "<<a<<' '<<b<<' '<<c<<endl;

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

#define ll long long

#define clr(a) memset(a,0,sizeof(a))

#define MOD 10000000000

#define INF 10000000000000



using namespace std;



int n,cp;

ll cost[MAX];

ll weight[MAX];

ll dp[MAX];





int main()

{

    cin>>n>>cp;

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

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

    }

    for(int i=0;i<=MAX;i++) dp[i]=INF;

    dp[0]=0;

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

        for(int j=MAX;j>0;j--){

            if(j-cost[i]>=0 && dp[j-cost[i]]!=INF){

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

            }

        }

    }

    for(int j=MAX;j>=0;j--){

        if(dp[j]<=cp){

            cout<<j<<endl;

            break;

            //break;

        }

    }



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.7368633632669348x):
#include<bits/stdc++.h>

using namespace::std;



const int MAX = 100000 + 5;

const int inf = 1e9 + 10;



int n;

int C;

int memo[MAX];



int main(){

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

	int sum = 0;

	for(int i = 1; i < MAX; i++) memo[i] = inf;

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

		int w, v;

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

		sum += v;

		for(int j = sum; j >= v; j--){

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

		}

	}

	int ans = 0;

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

		if(memo[i] <= C) ans = i;

	}

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

	return 0;

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



using namespace std;



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

#define allr(v)				((v).rbegin()), ((v).rend())

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

#define ll long long

#define pb					push_back

#define fIO ios_base::sync_with_stdio(0);cin.tie(0);



const ll oo = 1e8;

const double pi = 3.1415926535897;





const double EPS = (1e-7);

int dcmp(double x, double y) {	return fabs(x-y) <= EPS ? 0 : x < y ? -1 : 1;	}





typedef vector<int>                vi;

typedef vector<pair<int,int>>      vii;

typedef vector<ll>                 vll;

typedef vector<double>             vd;

typedef vector< vi >               vvi;

typedef vector< vd >               vvd;

typedef vector<string>             vs;

typedef pair<int, int>             pii;



/**

    * PLEASEEE READDD PROBLEEEM CARFULLY ****

    * Always check for OVERFLOW ll vs int

    * Always check for array bounds

    * Check for function return

    * READ OUTPUT SECTION AGAIN!!!

**/



struct edge{

    int from, to, w;



    edge(){}

    edge(int x, int y, int we): from(x), to(y), w(we){}



    bool operator < (const edge &rhs) const{

        return w > rhs.w;

    }



};



const int N = 100 + 5, V = 1e5 + 5 ;



int dp[V];

int w[N], v[N];

int n, ans = 0;

//

//int sol(int i, int vS, int wS) {

//    if(i == n) {

//        ans = max(ans, vS);

//        return 0;

//    }

//

//    int &ret = dp[i][vS];

//    if(ret != -1) return ret;

//

//    ret = sol(i + 1, vS, wS);

//

//    if(wS - w[i] >= 0) ret = max(ret, sol(i + 1, vS + v[i], wS - w[i]));

//

//    return ret;

//}



int main(){

    fIO



//    freopen("", "r", stdin);

//    freopen("", "w", stdout);



    fill(dp, dp + V, 1e9 + 5);



    int cap;

    cin >> n >> cap;



    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 = V - 5; j >= 1; j--) {

            if(j >= v[i]) {

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

            }

        }

    }





    for(int i = V - 5; i >= 1; i--) {

        if(dp[i] <= cap) {

            cout << i << "\n";

            break;

        }

    }

}
















// Faster C++ solution for the same problem (measured speedup: 1.6167449254590456x):
#include<bits/stdc++.h>

using namespace::std;



const int MAX = 100000 + 5;

const int inf = 1e9 + 10;



int n;

int C;

int memo[MAX];



int main(){

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

	int sum = 0;

	for(int i = 1; i < MAX; i++) memo[i] = inf;

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

		int w, v;

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

		sum += v;

		for(int j = sum; j >= v; j--){

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

		}

	}

	int ans = 0;

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

		if(memo[i] <= C) ans = i;

	}

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

	return 0;

}
// Slower C++ solution:
#pragma region

#include <algorithm>

#include <cstring>

#include <iostream>

#include <map>

#include <queue>

#include <stack>

#include <string>

#include <vector>

typedef long long ll;

#define tr t[root]

#define lson t[root << 1]

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

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

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

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

template <typename T>

void read(T &x) {

    x = 0;

    char ch = getchar();

    ll f = 1;

    while (!isdigit(ch)) {

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

        ch = getchar();

    }

    while (isdigit(ch)) {

        x = x * 10 + ch - 48;

        ch = getchar();

    }

    x *= f;

}

using namespace std;

/*

 *                                                     __----~~~~~~~~~~~------___

 *                                    .  .   ~~//====......          __--~ ~~

 *                    -.            \_|//     |||\\  ~~~~~~::::... /~

 *                 ___-==_       _-~o~  \/    |||  \\            _/~~-

 *         __---~~~.==~||\=_    -_--~/_-~|-   |\\   \\        _/~

 *     _-~~     .=~    |  \\-_    '-~7  /-   /  ||    \      /

 *   .~       .~       |   \\ -_    /  /-   /   ||      \   /

 *  /  ____  /         |     \\ ~-_/  /|- _/   .||       \ /

 *  |~~    ~~|--~~~~--_ \     ~==-/   | \~--===~~        .\

 *           '         ~-|      /|    |-~\~~       __--~~

 *                       |-~~-_/ |    |   ~\_   _-~            /\

 *                            /  \     \__   \/~                \__

 *                        _--~ _/ | .-~~____--~-/                  ~~==.

 *                       ((->/~   '.|||' -_|    ~~-/ ,              . _||

 *                                  -_     ~\      ~~---l__i__i__i--~~_/

 *                                  _-~-__   ~)  \--______________--~~

 *                                //.-~~~-~_--~- |-------~~~~~~~~

 *                                       //.-~~~--\

 *                       ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

 */



#pragma endregion

const int maxn = 1e5 + 5;

ll n, W, w[maxn], val[maxn], dp[maxn];

int main() {

    IO;

    cin >> n >> W;

    rep(i, 1, n) cin >> w[i] >> val[i];

    rep(i, 1, 1e5) dp[i] = 0x3f3f3f3f;

    dp[0] = 0;

    rep(i, 1, n) for (int t = 1e5; t >= val[i]; --t) dp[t] = min(dp[t], dp[t - val[i]] + w[i]);

    for (int i = 1e5; i; --i)

        if (dp[i] <= W) {

            cout << i << endl;

            return 0;

        }

}

// Faster C++ solution for the same problem (measured speedup: 1.8354188538756286x):
#include<bits/stdc++.h>

using namespace::std;



const int MAX = 100000 + 5;

const int inf = 1e9 + 10;



int n;

int C;

int memo[MAX];



int main(){

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

	int sum = 0;

	for(int i = 1; i < MAX; i++) memo[i] = inf;

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

		int w, v;

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

		sum += v;

		for(int j = sum; j >= v; j--){

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

		}

	}

	int ans = 0;

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

		if(memo[i] <= C) ans = i;

	}

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

	return 0;

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

#define ll long long

using namespace std;

#define inf 0x3f3f3f3f 

#define MAXN 1019191130

int main(){

	ll  n , W; cin >> n >> W;

    ll w[101] , v[101] , dp[100001];

    for(int i = 0 ; i <= 100000; ++i) dp[i] = 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 =  100000 ; j >= v[i] ; j--){

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

        }

    }

    int j = 100001;

    while(j-- && dp[j] > W);



    cout<< j <<endl;

}

// Faster C++ solution for the same problem (measured speedup: 1.8678643934168226x):
#include<bits/stdc++.h>

using namespace::std;



const int MAX = 100000 + 5;

const int inf = 1e9 + 10;



int n;

int C;

int memo[MAX];



int main(){

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

	int sum = 0;

	for(int i = 1; i < MAX; i++) memo[i] = inf;

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

		int w, v;

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

		sum += v;

		for(int j = sum; j >= v; j--){

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

		}

	}

	int ans = 0;

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

		if(memo[i] <= C) ans = i;

	}

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

	return 0;

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

using namespace std;

int main()

{

	 int n,W;

    cin>>n>>W;

    long long w[n];

    long long v[n];

    for(int i=0;i<n;i++) cin>>w[i]>>v[i];

    long long dp[100005];

    for(int i=1;i<100005;i++)

    {

    	dp[i]=1e15;

	}

    dp[0]=0;

    int res=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]<=W){

                res=max(res,j);

            }

        }

    }

    cout<<res<<endl;

}

// Faster C++ solution for the same problem (measured speedup: 2.0694661684944067x):
#include<bits/stdc++.h>

using namespace::std;



const int MAX = 100000 + 5;

const int inf = 1e9 + 10;



int n;

int C;

int memo[MAX];



int main(){

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

	int sum = 0;

	for(int i = 1; i < MAX; i++) memo[i] = inf;

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

		int w, v;

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

		sum += v;

		for(int j = sum; j >= v; j--){

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

		}

	}

	int ans = 0;

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

		if(memo[i] <= C) ans = i;

	}

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

	return 0;

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

#include <iostream>

using namespace std;

typedef long long ll;

const int MAXN=1e5+7;

int N,W,ans;

ll w[MAXN],v[MAXN],f[MAXN];

int main()

{

	cin>>N>>W;

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

	memset(f,127,sizeof(f));

	f[0]=0;

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

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

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

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

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

	cout<<ans<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.8489338299560412x):
#include<bits/stdc++.h>

using namespace::std;



const int MAX = 100000 + 5;

const int inf = 1e9 + 10;



int n;

int C;

int memo[MAX];



int main(){

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

	int sum = 0;

	for(int i = 1; i < MAX; i++) memo[i] = inf;

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

		int w, v;

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

		sum += v;

		for(int j = sum; j >= v; j--){

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

		}

	}

	int ans = 0;

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

		if(memo[i] <= C) ans = i;

	}

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

	return 0;

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

#include <cassert>

#include <algorithm>

#define N int(1e5+1)

using namespace std;

typedef pair<int,int> pii;

int dp[N];

struct{

    int w,v;

}a[100];

int main(int argc,char ** argv)

{

    int n,w;

    cin>>n>>w;

    fill(dp+1,dp+N,w+1);

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

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

    }

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

        for(int j=N-1;j>=a[i].v;--j){

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

        }

    }

    int i;

    for(i=N-1;dp[i]>w;--i);

    cout<<i<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.3651612397444721x):
#include<bits/stdc++.h>

using namespace::std;



const int MAX = 100000 + 5;

const int inf = 1e9 + 10;



int n;

int C;

int memo[MAX];



int main(){

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

	int sum = 0;

	for(int i = 1; i < MAX; i++) memo[i] = inf;

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

		int w, v;

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

		sum += v;

		for(int j = sum; j >= v; j--){

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

		}

	}

	int ans = 0;

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

		if(memo[i] <= C) ans = i;

	}

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

	return 0;

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

 

#define f(i,a,b) for( ll i = a; i < b ; i++ )

#define af(i,a,b) for( ll i = a; i >= b ; i--)

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

#define arep(i,a,b,k) for( ll i = a; i >= b ; i-= k)

#define ones(x) (ll) __builtin_popcount(x)

#define fs first

#define sc second

#define pb push_back

#define po pop_back

#define mp make_pair  

#define sz(a) (ll) a.size()

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

#define sor(a) sort( a.begin(), a.end() )

#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL)

#define ller ios::sync_with_stdio(false);cin.tsie(nullptr);cout.tie(nullptr)

#define watch(x) cout << (#x) << " is #" << (x) <<"#" <<endl

#define PI 3.1415926535

using namespace std;

typedef long long ll;

typedef long double ld; 

typedef pair<ll,ll> ii ;

typedef vector<ll>  vi ;

typedef vector<ii> vii ;

const ll mod = 1e9+7;

const ll MAX = 1e5+5;

const ll inf = 1e18+7;



ll dp[MAX];

int main() {

  fastio;

  ll n,capac;

  cin>>n>>capac;

  f(i,0,MAX+1) dp[i]=inf;

  dp[0]=0;

  f(i,0,n){

    ll w,v;

    cin>>w>>v;

    af(j,MAX,v){

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

    }

  }

  af(j,MAX,0){

  if(dp[j]<=capac){cout<<j<<"\n";break;}

  }

  return 0; 

}

// Faster C++ solution for the same problem (measured speedup: 1.829760155477563x):
#include<bits/stdc++.h>

using namespace::std;



const int MAX = 100000 + 5;

const int inf = 1e9 + 10;



int n;

int C;

int memo[MAX];



int main(){

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

	int sum = 0;

	for(int i = 1; i < MAX; i++) memo[i] = inf;

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

		int w, v;

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

		sum += v;

		for(int j = sum; j >= v; j--){

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

		}

	}

	int ans = 0;

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

		if(memo[i] <= C) ans = i;

	}

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

	return 0;

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

//#pragma GCC optimize("Ofast")

//#pragma GCC target("avx,avx2,fma")



using namespace std;



typedef long long ll;



//#include <ext/pb_ds/assoc_container.hpp>

//#include <ext/pb_ds/tree_policy.hpp>

//using namespace __gnu_pbds;

//typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>pb_ds;



#define IN freopen("in.txt","r",stdin)

#define OUT freopen("out.txt","w",stdout)

#define pi pair<int,int>

#define F first

#define S second

#define pb push_back

#define pp pop_back

#define mod 1000000007

#define pai acos(-1)

#define N 100000



ll Weight[N+5],Profit[N+5],dp[2][N+5];



int main ()

{

   ios_base::sync_with_stdio(false);

   cin.tie(NULL);

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

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

      cin>>Weight[i]>>Profit[i];

   }

   fill(dp[0],dp[0]+2*(N+5),INT_MAX*105);

   dp[(n+1)%2][0]=0;

   for(int i=n;i>=1;i--){

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

         ll op1=dp[(i+1)%2][p];

         ll op2=(p-Profit[i]>=0)?Weight[i]+dp[(i+1)%2][p-Profit[i]]:INT_MAX*105;

         dp[i%2][p]=min(op1,op2);

      }

   }

   int ans=0;

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

      if(dp[1][i]<=w){

         ans=max(ans,i);

      }

   }

   cout<<ans<<"\n";

   return 0;

}




// Faster C++ solution for the same problem (measured speedup: 2.6707040293148734x):
#include<bits/stdc++.h>

using namespace::std;



const int MAX = 100000 + 5;

const int inf = 1e9 + 10;



int n;

int C;

int memo[MAX];



int main(){

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

	int sum = 0;

	for(int i = 1; i < MAX; i++) memo[i] = inf;

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

		int w, v;

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

		sum += v;

		for(int j = sum; j >= v; j--){

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

		}

	}

	int ans = 0;

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

		if(memo[i] <= C) ans = i;

	}

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

	return 0;

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

using namespace std;

typedef long long int ll;

const int MAX=1e5+10;

ll dp[MAX];

int main (){

    int N,W;cin >> N >> W;

    memset(dp,63,sizeof dp);

    dp[0]=0;

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

        int w,val;

        cin >> w >> val;

        for(int j=MAX;j>=0;j--){

            if(j-val>=0)

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

        }



    }

    int res=0;

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

        if(dp[i]<=W) res = max(res,i);

    }

    cout<<res<<endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 3.019735942024552x):
#include<bits/stdc++.h>

using namespace std;

using ll=long long;

const int N=1e5+10,M=1e2+10;

const int mod=1e9+7;

const int inf=1e9;

int w[M],v[M];

int dp[M][N];

 

int main(){

    int W,n;

    int val=0;

    cin>>n>>W;

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

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

        val+=v[i];

    }

    dp[0][0]=0;

    for(int i=1;i<=val;i++) dp[0][i]=inf;

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

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

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

            if(i>=v[j]){

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

            }

        }

    }

    

    int ans=0;

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

        if(dp[n][i]<=W) ans=i;

    }

    

    cout<<ans<<'\n';

    return 0;

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

#include <iostream>

#include <string>

#include <vector>

#include <algorithm>

#include <sstream>

#include <queue>

#include <deque>

#include <bitset>

#include <iterator>

#include <list>

#include <stack>

#include <map>

#include <set>

#include <functional>

#include <numeric>

#include <utility>



#include <cmath>

#include <iomanip>



using namespace std;



int main()

{

  ios::sync_with_stdio(false);

  cin.tie(0);

  int n;

  cin >> n;

  int w;

  cin >> w;

  int weights[n];

  long long int values[n];

  long long int cur;

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

    cin >> cur;

    weights[i] = cur;

    cin >> cur;

    values[i] = cur;

  }

  long long int dp[n+1][(1000*n)+1];

  //dp[n][v] represents the minimum weight needed to get value v and only using items with number up to n.

  for (int i = 0; i < n+1; i++) {

    dp[i][0] = 0;

    for (int j = 1; j < (1000*n)+1; j++) {

      dp[i][j] = w+1;

    }

  }

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

    for (int wanted = 0; wanted < (1000*n); wanted++) {

      dp[minInd+1][wanted] = min(dp[minInd+1][wanted], dp[minInd][wanted]);

      if (dp[minInd][wanted]+weights[minInd] <= w) {

	dp[minInd+1][wanted+values[minInd]] = min(dp[minInd+1][wanted+values[minInd]], dp[minInd][wanted]+weights[minInd]);

      }

    }

  }

  for (int i = (1000*n); i > 0; i--) {

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

      cout << i << "\n";

      break;

    }

  }

  return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.2275177308011824x):
#include<bits/stdc++.h>

using namespace std;

using ll=long long;

const int N=1e5+10,M=1e2+10;

const int mod=1e9+7;

const int inf=1e9;

int w[M],v[M];

int dp[M][N];

 

int main(){

    int W,n;

    int val=0;

    cin>>n>>W;

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

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

        val+=v[i];

    }

    dp[0][0]=0;

    for(int i=1;i<=val;i++) dp[0][i]=inf;

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

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

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

            if(i>=v[j]){

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

            }

        }

    }

    

    int ans=0;

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

        if(dp[n][i]<=W) ans=i;

    }

    

    cout<<ans<<'\n';

    return 0;

}