## 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 "+".

• commented on March 17, 2023, 1:37 p.m.

.

• commented on March 17, 2023, 1:40 p.m.

+1 ban vote

• commented on March 17, 2023, 1:38 p.m.

+1 ban vote

• commented on Feb. 11, 2023, 8:01 a.m.

nhờ bài lày mà mình đc 0,01 điểm.cảm ơn VNOI

• commented on Jan. 21, 2023, 2:36 a.m. edit 2

valĩd

### include <iostream>

using namespace std;

int main() { int a, b; cin >> a >> b;

int ans = 0;

for (int i = 0; i < 1000000000; >!i++) { ans = a + b; }

cout << ans << endl; return 0; }

• commented on Dec. 28, 2022, 3:41 a.m.

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• commented on Dec. 26, 2022, 12:28 a.m. edited

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• commented on Dec. 25, 2022, 1:57 a.m.

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• commented on Sept. 20, 2022, 3:19 p.m.

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• commented on Sept. 22, 2022, 10:29 a.m.

Bài này cho 2 số A và B trên cùng một dòng. Hàm input() thì đọc hết nguyên một dòng luôn.

Bạn có thể đọc input như sau:

a, b = map(int, input().split())

• commented on Aug. 8, 2022, 12:17 p.m.

xin 1 upvote

• commented on Aug. 8, 2022, 3:17 p.m.

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• commented on July 29, 2022, 10:11 a.m.

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• commented on March 15, 2023, 2:19 p.m.

hay

• commented on June 10, 2022, 9:28 a.m.

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• commented on June 9, 2022, 2:09 a.m. edit 2

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• commented on Sept. 8, 2021, 9:36 a.m.

uoc j

• commented on Sept. 8, 2021, 9:37 a.m.

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• commented on Aug. 5, 2021, 3:19 a.m.

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; }

• commented on Dec. 21, 2022, 8:50 a.m. edited

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