hdoj_Problem1.1.8_A+B for Input-Output Practice (VIII)

A+B for Input-Output Practice (VIII)

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/32768 K (Java/Others)

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.

留意between

Sample Input

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

Sample Output

10

15

6

• 第一次提交没有注意到between，OJ返回Presentation Error

between的意思是，输出的数字之间用空行分开，但最后一个输出下面是没有空行

• 增加判断条件之后第二次提交通过。

Solution

#include <cstdio>
using namespace std;
#define DEBUG 0	//本地调试定义为1, 提交时定义为0

#if DEBUG
#else
#define fp stdin	//当DEBUG被定义为0时, fp被替换为stdin
#endif

int main() {
#if DEBUG
	FILE* fp;
	fp = fopen("a.txt", "r");	//a.txt作为文件输入
#endif
	int n;
	while (fscanf(fp, "%d", &n) != EOF) {
		break;
	}
	int* sum_arr = new int[n];
	for (int i = 0; i < n; i++) {
		int m;
		while (fscanf(fp, "%d", &m) != EOF) {
			break;
		}
		int sum = 0;
		for (int j = 0; j < m; j++) {
			int add;
			while (fscanf(fp, "%d", &add) != EOF) {
				break;
			}
			sum += add;
		}
		
		sum_arr[i] = sum;
	}
	for (int i = 0; i < n; i++) {
		printf("%d\n", sum_arr[i]);
		if (i < n - 1) {
			printf("\n");
		}
	}
	delete[] sum_arr;
	return 0;
}

如果觉得要反复修改宏定义的DEBUG麻烦，可以把调试语句改为

#ifdef DEBUG
#else
#define fp stdin
#endif

int main() {
#ifdef DEBUG
	FILE* fp;
	fp = fopen("a.txt", "r");
#endif

需要进行本地调试时，使用g++编译选项定义DEBUG

$ g++ 1.1.8.cpp -DDEBUG

运行

$./a