1.1.6和1.1.4相似,不过是少了n=0时不做处理那个条件,1.1.7和1.1.1相似,不过多了输出一个空行而已。1.1.8和1.1.5类似,下面来看看1.1.8。
题目描述:
Problem Description
Your task is to calculate the sum of some integers.
Input
Input contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line.
Output
For each group of input integers you should output their sum in one line, and you must note that there is a blank line between outputs.
Sample Input
3
4 1 2 3 4
5 1 2 3 4 5
3 1 2 3
Sample Output
10
15
6
题意:先在第一行输入一个数字N,表示你将处理N行的数据,接下来输入M,表示改行要处理的M个数据,然后求和,最后要求在两个输出的和之间要有一个空行。(注意:是两个输出结果之间)N行就是N个M。
上代码:
#includeint main() { int N, M, i, j, a; int sum = 0; while (scanf("%d", &N) != EOF) { for (i = 0; i < N; i++) { scanf("%d", &M); for (j = 0; j < M; j++) { scanf("%d", &a); sum += a; } if (i != N - 1) printf("%d\n\n", sum); else printf("%d\n", sum); sum = 0; } } return 0; }
难点:输出结果后再输出空行,但是输出到最后一行时,将不再输出空行。