题目1143：Primary Arithmetic

最新推荐文章于 2024-01-07 11:02:09 发布

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最新推荐文章于 2024-01-07 11:02:09 发布

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本文介绍了一个程序设计问题，即计算两个多位数相加时产生的进位操作次数。通过从右向左逐位相加的方式，该程序帮助教育工作者评估这类数学题目的难度。

题目描述：

Children are taught to add multi-digit numbers from right-to-left one digit at a time. Many find the "carry" operation - in which a 1 is carried from one digit position to be added to the next - to be a significant challenge. Your job is to count the number of carry operations for each of a set of addition problems so that educators may assess their difficulty.

输入：

Each line of input contains two unsigned integers less than 10 digits. The last line of input contains 0 0.

输出：

For each line of input except the last you should compute and print the number of carry operations that would result from adding the two numbers, in the format shown below.

样例输入：

样例输出：

NO carry operation.
3 carry operations.
1 carry operation.

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;

char op1[11];
char op2[11];
int main() {
	while (cin >> op1 >> op2) {
		if (strlen(op1) == 1 && strlen(op2) == 1 && op1[0] == '0' && op2[0]
				== '0')
			break;
		int pos1 = strlen(op1) - 1;
		int pos2 = strlen(op2) - 1;
		int sum = 0, carry = 0, cnt = 0;
		while (pos1 >= 0 && pos2 >= 0) {
			sum = (op1[pos1] - '0') + (op2[pos2] - '0') + carry;
			carry = sum / 10;
			if (sum >= 10)
				cnt++;
			pos1--;
			pos2--;
		}
		while (pos1 >= 0) {
			sum = (op1[pos1] - '0') + carry;
			carry = sum / 10;
			if (sum >= 10)
				cnt++;
			pos1--;
		}
		while (pos2 >= 0) {
			sum = (op2[pos2] - '0') + carry;
			carry = sum / 10;
			if (sum >= 10)
				cnt++;
			pos2--;
		}
		if (cnt == 0)
			printf("NO carry operation.\n");
		else if (cnt == 1)
			printf("1 carry operation.\n");
		else
			printf("%d carry operations.\n", cnt);

	}
	return 0;
}

题目链接：

http://ac.jobdu.com/problem.php?pid=1143