A+B 三连_sum+b-优快云博客

本文链接：https://blog.youkuaiyun.com/llipoli/article/details/80317580

本文介绍了三个不同难度级别的A+B问题，分别涉及基本的加法运算、多组输入处理及复杂的数据读取与处理技巧。通过这些例子，读者可以了解如何在C/C++中处理不同的输入输出格式，并学会解决常见编程竞赛中的基础算术问题。

刚入门的时候肯定都要做一些简单的题目，一是为了适应，二是为了增强信心。

A+B（一）

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 215 Solved: 143

Description

求两个整数的和并输出

Input

第一行输入一个整数n（1 <= n <= 100）
接着输入n组数，每组两个整数

Output

输出每组整数的和并输出

Sample Input

Sample Output

4
7
33

#include"stdio.h"

int main ()

{

    int n, b, a;

    int c;

    while(~scanf("%d\n",&n))

    {   c=0;

        for(int i=0;i<n;i++)

{

            scanf("%d %d",&a,&b);

            c=a+b;

            printf("%d\n",c);

}

}

    return 0;

}

A+B(二)

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 200 Solved: 130

Description

Your task is to calculate the sum of some integers.

Input

Input contains multiple test cases, and one case one line.

Each case starts with an integer N(1 <= N <= 10000), and then N integers Ai( |Ai| <= 10000) follow in the same line.

Output

For each test case you should output the sum of N integers in one line, and with one line of output for each line in input.

Sample Input

4 1 2 3 4
5 1 2 3 4 5

Sample Output

10
15

#include <iostream>
using namespace std;
int main()
{
int n,i,a,b;
while(scanf("%d",&n)==1)
{
a = 0;
for(i=0;i<n;i++)
{
scanf("%d",&b);
a=a+b;
}
printf("%d\n",a);
}
return 0;
}

A+B （三）

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 117 Solved: 79

Description

Your task is to calculate the sum of some integers.

Input

Input contains an integer N(0 < N <= 100) in the first line, and then N lines follow. Each line starts with a integer M, and then M(0 < M <= 100) integers follow in the same line.

Output

For each group of input integers you should output their sum in one line, and with one line of output for each line in input.

Sample Input

2
4 1 2 3 4
5 1 2 3 4 5

Sample Output

10
15

#include<stdio.h>

int main()

{

    int a, i ,m;

    while(~scanf("%d",&m))

{

        int b; 

        for(i=0;i<m;i++)

{

            int sum=0;

            int il;

            scanf("%d",&a);

            for(il=1;il<=a;il++)

{

                scanf("%d",&b);

                sum=sum+b;

}

            printf("%d\n",sum);

}

}

    return 0;

}