POJ 2562 Primary Arithmetic（我的水题之路——模拟加法进位）

Primary Arithmetic

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8450 Accepted: 3166

Description

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 456555 555123 5940 0

Sample Output

No carry operation.3 carry operations.1 carry operation.

Source

Waterloo local 2000.09.23

又是一道英文题。给两个不超过十位的正整数，问它们相加的过程中，一共有几次进位。

高精度模拟，做过高精度加法的同学都不会觉得太难吧。

注意点：

1）当仅有一次进位的时候，输出"operation"，其他时候输出"operations".

2）进位时候，当两个数字长度不相同时，考虑多出部分的进位，如：“99999 + 1”。

代码（1AC）：

#include <cstdio>#include <cstdlib>#include <cstring>char num1[20];char num2[20];int main(void){    int add, tmp;    int carry;    int i, j;    int len1, len2;    while (scanf("%s %s", num1, num2), strcmp(num1, "0") != 0 || strcmp(num2, "0") != 0){        getchar();        len1 = strlen(num1);        len2 = strlen(num2);        for (carry = add = 0, i = len1 - 1, j = len2 - 1 ; i >= 0 && j >= 0; i--, j--){            if ((num1[i] - '0') + (num2[j] - '0') + add >= 10){                add = 1;                carry ++;            }            else{                add = 0;            }        }        for (; i >= 0; i--){            if (num1[i] - '0' + add >= 10){                carry++;                add = 1;            }            else {                add = 0;            }        }        for (; j >= 0; j--){            if (num2[j] - '0' + add >= 10){                carry++;                add = 1;            }            else {                add = 0;            }        }        if (carry == 0){            printf("No carry operation.\n");        }        else if (carry == 1){            printf("1 carry operation.\n");        }        else{            printf("%d carry operations.\n", carry);        }    }    return 0;}