A+B for Input-Output Practice (VI)
题目描述:
Your task is to calculate the sum of some integers.
输入格式:
Input contains multiple test cases, and one case one line. Each case starts with an integer N, and then N integers follow in the same line.
输出格式:
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.
接下来我们继续看1090题,A + B VI 这一题也是多组输入,如输入:
4 1 2 3 4
5 1 2 3 4 5
则输出:
10
15
但并没有说明什么时候结束,因此需要在循环中不间断的接收,后面的输入就没有难度了,依旧n--循环依次接收,记得累加求和的变量每次都归零就OK了
参考C语言代码:
#include<stdio.h>
int main()
{
int n;
int a,sum=0;
while(scanf("%d",&n)!=EOF)
{
sum=0;
while(n--)
{
scanf("%d",&a);
sum=sum+a;
}
printf("%d\n",sum);
}
return 0;
}
参考C++代码:
#include<iostream>
using namespace std;
int main()
{
int n;
int a,sum=0;
while(cin>>n)
{
sum=0;
while(n--)
{
cin>>a;
sum=sum+a;
}
cout<<sum<<endl;
}
return 0;
}
参考Java代码:
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()) {
int sum=0;
int n = scanner.nextInt();
while (n--!=0){
int a=scanner.nextInt();
sum=sum+a;
}
System.out.println(sum);
}
}
}
参考Python代码:
while True:
try:
sum=0
a=list(map(int,input().split()))
for i in range(1,len(a)):
sum+=a[i]
print(sum)
except:
break
以上就是这期的内容了 有计算机志同道合的同学可以私信我 到时候大家一起鼓励联系 提高自身能力
标签:scanner,int,sum,输入输出,ACM,蓝桥,--,while,line From: https://blog.csdn.net/xinfanyyds/article/details/143217324