swjtu 1962 A+B(模拟)

题目链接：http://acm.swjtu.edu.cn/JudgeOnline/showproblem?problem_id=1962

问题思路：考察编程基础的问题，涉及到字符串转为数字的问题。

代码如下：

#include <stdio.h>
#include <string.h>

#define MAX_N 1000

int num_a[MAX_N], num_b[MAX_N];
int digit_a, digit_b;
char character[5][MAX_N];
char example[5][40] =
{
    "+-+  ++-++-++ ++-++-++-++-++-+",
    "| |  |  |  || ||  |    || || |",
    "+ +  ++-++-++-++-++-+  ++-++-+",
    "| |  ||    |  |  || |  || |  |",
    "+-+  ++-++-+  ++-++-+  ++-++-+",
};

int CharToNum(int char_i, int exap_i)
{
    int c_i = char_i;
    int e_j = exap_i;

    if (character[0][c_i] == ' ' &&
        character[0][c_i + 1] == ' ' &&
        character[0][c_i + 2] == ' ')
        return -1;

    for (int i = 0; i < 5; ++i)
    {
        for (int j = 0; j < 3; ++j)
        {
            if (character[i][c_i + j] !=
                example[i][e_j + j])
                return 0;
        }
    }

    return 1;
}

void output(int ans)
{
    int digit = 0;
    int num[MAX_N];
    int divide = ans;
    int rest = 0;

    memset(num, 0, sizeof(num));
    for (int i = 0; divide != 0 || rest != 0; ++i)
    {
        num[i] = divide % 10;
        divide /= 10;
        digit = i;
    }

    for (int i = 0; i < 5; ++i)
    {
        for (int j = digit; j >= 0; --j)
        {
            int num_line = num[j] * 3;

            for (int k = 0; k < 3; ++k)
                printf("%c", example[i][num_line + k]);

            if (j == 0)
                printf("
");
            else
                printf(" ");
        }
    }
}

int main()
{
    int len = 0;
    int flag = false;

    for (int i = 0; i < 5; ++i)
        gets(character[i]);

    len = strlen(character[0]);

    for (int i = 0; i < len; i += 4)
    {
        for (int j = 0; j < 30; j += 3)
        {
            int result = CharToNum(i, j);

            if (result == -1)
            {
                flag = true;
                break;
            }

            if (!flag && result != 0)
            {
                num_a[digit_a++] = j / 3;
                break;
            }
            else
            if (flag && result != 0 && result != -1)
            {
                num_b[digit_b++] = j / 3;
                break;
            }
        }
    }

    int a = 0, b = 0;

    for (int i = 0; i < digit_a; ++i)
        a = a * 10 + num_a[i];
    for (int i = 0; i < digit_b; ++i)
        b = b * 10 + num_b[i];

    output(a + b);

    return 0;
}