#include <bits/stdc++.h> using namespace std; int a,b,x[4],y[4],bz; int main() { cin>>a>>b; for(int i=1;i<=3;i++) { cin>>x[i]>>y[i]; if(b/y[i]*x[i]>=a) { bz=1; cout<<"YES"; return 0; } } if(bz==0) { cout<<"NO"; } }

#include<iostream> using namespace std; int main(){ int a,b,x1,y1,x2,y2,x3,y3; cin>>a>>b>>x1>>y1>>x2>>y2>>x3>>y3; if(b/y1x1>=a||b/y2x2>=a||b/y3*x3>=a) cout<<"YES"; else cout<<"NO"; return 0; }

#include <bits/stdc++.h> using namespace std; int main() { int a,b,x1,y1,x2,y2,x3,y3; cin>>a>>b; }

#include <bits/stdc++.h> using namespace std; int main(){ int a,b,x1,x2,x3,y1,y2,y3,n; cin >> a >> b; cin >> x1 >> y1; cin >> x2 >> y2; cin >> x3 >> y3; if ((b / y1) * x1 >= a){ cout << "YES"; } else if ((b / y2) * x2 >= a){ cout << "YES"; } else if ((b / y3) * x3 >= a){ cout << "YES"; } else { cout << "NO"; } return 0; }

a,b=input().split() a=int(a) b=int(b) x,y=input().split() x=int(x) y=int(y) p,o=input().split() p=int(p) o=int(o) c,v=input().split() c=int(c) v=int(v) if b//yx>=a or b//op>=a or b//v*c>=a: print('YES') else: print('NO')

#include<iostream> using namespace std; int main(){ }

#include<iostream> using namespace std; int main(){ int a,b,x1,y1,x2,y2,x3,y3; cin>>a>>b; cin>>x1>>y1; cin>>x2>>y2; cin>>x3>>y3; if(b/y1x1>=a)cout<<"YES"; else if(b/y2x2>=a)cout<<"YES"; else if(b/y3*x3>=a)cout<<"YES"; else cout<<"NO"; return 0; }

#include<iostream> using namespace std; int main(){ int a, b, food[3][2], times; bool flag; cin>>a>>b; for (int i=0; i<3; i++){ for (int j=0; j<2; j++){ cin>>food[i][j]; } } for (int i=0; i<3; i++){ if (food[i][1]<=b){ if (a%food[i][0]!=0){ times=a/food[i][0]+1; }

#include <bits/stdc++.h> using namespace std; int main(){ int a,b; cin>>a>>b; int x1,y1,x2,y2,x3,y3; cin>>x1>>y1>>x2>>y2>>x3>>y3; int cnt=b/y1; int ans=x1cnt; if(a<=ans){ cout<<"YES"; return 0; } cnt=b/y2; ans=x2cnt; if(a<=ans){ cout<<"YES"; return 0; } cnt=b/y3; ans=x3*cnt; if(a<=ans){ cout<<"YES"; return 0; } }

这里也是写了两种解法啊！！！ First One 极致格式+不怕aac Second One 2. cstdio格式 First One Second One 第一次写题解，不要喷我。