A cộng B

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Points: 0.01 (partial)
Time limit: 1.0s
Memory limit: 512M
Input: stdin
Output: stdout

Problem source:
Just to test
Problem type
Allowed languages
C, C++, Java, Kotlin, Pascal, PyPy, Python, 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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  • -7
    caubechepcode   commented on 9, Feb, 2022, 15:38

    This comment is hidden due to too much negative feedback. Show it anyway.


  • -7
    nictysine1   commented on 8, Feb, 2022, 14:34

    This comment is hidden due to too much negative feedback. Show it anyway.


  • -15
    AnNoProo   commented on 30, Dec, 2021, 14:27

    This comment is hidden due to too much negative feedback. Show it anyway.


  • -6
    K24NVNTin   commented on 16, Nov, 2021, 15:57

    This comment is hidden due to too much negative feedback. Show it anyway.


  • -12
    K24LAQUAN   commented on 16, Nov, 2021, 15:57

    This comment is hidden due to too much negative feedback. Show it anyway.


  • -13
    mronjudge   commented on 30, Oct, 2021, 10:40

    This comment is hidden due to too much negative feedback. Show it anyway.


  • 7
    darkkcyan   commented on 8, Sep, 2021, 16:35

    thinkingcat


  • -28
    SPyofgame   commented on 8, Sep, 2021, 16:29

    This comment is hidden due to too much negative feedback. Show it anyway.


    • -4
      someone   commented on 9, Sep, 2021, 20:43

      chịu thật nếu idea như này cho mình xin code với


    • 7
      pichu   commented on 8, Sep, 2021, 16:36

      uoc j


  • 13
    caubecanlao   commented on 5, Aug, 2021, 10:19

    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; }


    • 0
      HN_CSP_FTRs   commented on 8, Jan, 2022, 20:05

      anh quả là hảo hán


    • -11
      nguyenvana123   commented on 5, Aug, 2021, 10:21

      This comment is hidden due to too much negative feedback. Show it anyway.