3 solutions
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0
#include <bits/stdc++.h> using namespace std; typedef long long LL; const int N = 100010; int a[N]; int b[N]; int SR(int x,int l,int r) { if(l >= r) return -1; while(l < r) { int mid = l + r + 1 >> 1; if(a[mid] <= x) { l = mid; } else r = mid - 1; } if(a[r]==x)return r; return -1; } int main () { int n ; cin >> n; for (int i = 1; i <= n; i ++) { cin >> a[i]; } int m; cin >> m; for (int i = 1; i <= m ; i ++) { cin >> b[i]; } for(int i = 1; i <= m; i ++) { cout << SR(b[i],0,n) <<' '; } return 0; }
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废话少说看代码
#include<bits/stdc++.h> using namespace std; const int N = 1e5 + 5; int num1[N]; int main(void) { int n, m; cin >> n; for (int i = 1; i <= n; i++) cin >> num1[i]; cin >> m; for (int i = 0; i < m; i++) { int t,l; cin >> t; l = upper_bound(num1 + 1, num1 + 1 + n, t) - num1-1; if (num1[l] != t) cout << -1 << " "; else cout << l << " "; } return 0; }
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#include<bits/stdc++.h> using namespace std; const int N=1e6+5; int f[N]; int n,l,r,m,g; int find(int k){ l=1,r=n; while(l<=r){ int mid=(l+r)/2; if(f[mid]<=k){ l=mid+1; } else{ r=mid-1; } } if(f[l-1]==k){ return l-1; } else{ return -1; } } int main(){ cin>>n; for(int i=1;i<=n;i++){ cin>>f[i]; } cin>>m; while(m--){ int g; cin>>g; cout<<find(g)<<" "; } return 0; }
- 1
Information
- ID
- 1231
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- 6
- Tags
- # Submissions
- 245
- Accepted
- 73
- Uploaded By