1001. A+B Format (20)

最新推荐文章于 2024-06-13 15:02:42 发布

Jack_zhuiyi

最新推荐文章于 2024-06-13 15:02:42 发布

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分类专栏：算法

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该博客介绍了一个计算两个整数a和b之和的问题，要求将结果以标准格式输出，即每三位数字之间用逗号分隔，除非总共有少于四个数字。

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时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).

Input

Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.

Output

For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.

Sample Input

-1000000 9

Sample Output

-999,991

#include <iostream>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <cmath>
#include <stack>
#include <string>
#include <queue>
#include <algorithm>
#include <cstdio>
using namespace std;
#define INF 0x3f3f3f3f
int main() {
	int a, b;
	stack<int> sta;
	while (~scanf("%d%d",&a,&b)) {
		int sum = a + b;
		if (sum < 0) {
			putchar('-');
			sum = abs(sum);
		}
		if (sum < 1000) {
			printf("%d\n", sum);
			continue;
		}
		while (sum) {
			int t = sum % 10;
			sum /= 10;
			sta.push(t);
		}
		bool flag = false;
		int s = ((int)sta.size()) % 3;
		for (int i = 0; i < s; ++i) {
			printf("%d", sta.top());
			sta.pop();
			flag = true;
		}
		if (flag) {
			putchar(',');
		}
		while (!sta.empty()) {
			printf("%d", sta.top());
			sta.pop();
			if (((int)sta.size()) % 3 == 0 && !sta.empty()) {
				putchar(',');
			}
		}
		putchar('\n');
	}
	return 0;
}

A+B,标准格式输出，注意绝对值<1000的情况，注意3的倍数位的情况。