10035 - Primary Arithmetic
Time limit: 3.000 seconds
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.
Input
Each line of input contains two unsigned integers less than 10 digits. The last line of input contains 0 0.Output
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.Sample Input
123 456 555 555 123 594 0 0
Sample Output
No carry operation. 3 carry operations. 1 carry operation.
有至少两个陷阱吧,详见注释。
完整代码:
/*0.019s*/
#include <cstdio>
int main(void)
{
int a, b;
int carry, c;///carry表示进位数
while (scanf("%d%d", &a, &b) , a + b)
{
carry = c = 0;
while (a || b) ///注意999+1的情况~
{
c = (a % 10 + b % 10 + c) > 9 ? 1 : 0;
carry += c;///或者写if(c) ++carry;
a /= 10;
b /= 10;
}
if (carry == 0)
puts("No carry operation.");
else if (carry == 1)
puts("1 carry operation.");
else
printf("%d carry operations.\n", carry);
}
return 0;
}

本文探讨了如何通过编程解决儿童数学教育中遇到的加法问题,特别是涉及多位数相加时的进位操作。文章提供了一个算法示例,用于计算任意两个小于10位数相加时的进位次数。
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