Submit solution
Points:
0.01 (partial)
Time limit:
1.0s
Memory limit:
512M
Input:
stdin
Output:
stdout
Problem source:
Problem type
Allowed languages
C, C++, Go, Java, Kotlin, Pascal, PyPy, Python, Rust, Scratch
Cho ~2~ số nguyên ~A~ và ~B~. Hãy tính ~A + B~.
Input
Gồm ~1~ dòng chứa ~2~ số nguyên ~A~ và ~B~ ~(1 \le A, B \le 1000)~, cách bởi ~1~ dấu cách.
Output
Ghi ra tổng ~A + B~.
Sample Input
3 4
Sample Output
7
Note
Gợi ý: Sử dụng toán tử "+".
Comments
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bài này phải giải bằng thuật toán đệ quy có nhớ, nhân ấn độ, lũy thừa nhị phân, quy hoạch động, dfs/bfs, eulertour, stack, nhân ma trận, kmp, hashing, segment tree, quick sort, BIT, chặt nhị phân, tarjan, LCA, bảng thưa, siêu cấp tổ hợp toán học, tourist
gspvh <3 gen
gspvh kh cuti
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xl moi nguoi minh bi troll :(
Mình cảm thấy khá khó chịu khi đọc vào phần bình luận của bài này các bạn sao cứ spam ấy nhỉ,và cũng đã có nhiều bài bị spam bình luận giống vậy ,hãy xây dựng một cộng đồng VNOI có văn hoá chứ đừng có như vậy.Xin cảm ơn(mình chỉ là coder thôi)
Thấy bài dễ mà nhiều ông vào chỉ v :))
Bài A + B làm thế này
Template Java input + output đơn giản có unit test (driver class):
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.
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Bài này chủ yếu để làm quen input, output, và run/submit với constraint thời gian thôi bạn.
(Về phía server, cũng tiết kiệm tài nguyên và giảm tỷ lệ nộp sai khi beginners submit các bài actual practice)
Tương tự SPOJ cũng có bài làm quen: https://www.spoj.com/problems/TEST , https://www.spoj.com/problems/J4FUN/
https://luyencode.net/problem/cb02 https://ctoj.net/problem/PA0004 https://lqdoj.edu.vn/problem/w01
https://lequydon.ntucoder.net/Problem/Details/1
https://tek4.vn/luyen-tap/1000-bai-tap-lap-trinh-c/post3-ai-cong-nhanh-nhat
http://csloj.ddns.net/problem/1
(CodeChef cũng có problems/TEST và problems/INTEST giống SPOJ)
đâu phải ai cũng làm được
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xam
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xin 1 upvote
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công thức quy hoạch động khá đơn giản
include <bits/stdc++.h>
using namespace std; const int N = 1e3 + 1e2; int dp[N][N]; int main() { cin.tie(0)->syncwithstdio(0); dp[1][0] = 1; dp[0][1] = 1; for (int i = 1; i <= 1000; i++) { for (int j = 1; j <= 1000; j++) { dp[i][j] = max({dp[i - 1][j] - (-1), dp[i][j - 1] - (-1), dp[i - 1][j - 1] - (-2)}); } } int a, b; cin >> a >> b; cout << max(dp[a][b], dp[b][a]); return 0; }
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