ZOJ_1178_Booklet Printing

Mon, 18 Oct 2010 09:05:00 GMT

ZOJ 1178 Booklet Printing

�q�道题是一道简单的模拟题，但是做�v来有点繁琐，不仅处理�q�程�J�琐�Q�输��Z��J�琐。这个问题大意是打印文章�Q�双面打华ͼ�一面打��C��，�l�定��|��Q�然后输出每��늺�打印的页码。要求这些纸攑֜�一��P��从中间折叠后�Q�页码连�l�递增。找两张�U怸�试就知道怎么回事�?ji��n)。用front[0]�?/span>front[1]分别表示正面打印的页码数�Q?/span>back[0]�?/span>back[1]表示背面打印的页码数。对�?/span>4��늚�文章�Q�毫无疑�?/span>front[1] = 1;back[0]=2;back[1]=3;front[0]=4。在输出的时�?/span>back与冒号之间有�I�格�Q��?/span>front与冒号之间无�I�格�Q�开始还?sh��)��?f��)是题目打错了(ji��n)�Q�自作聪明的该了(ji��n)下，提交��?/span>WA�?ji��n)，后来改过来一看，原来是多打个�I�格是因�?/span>front�?/span>back多个字母�Q��ؓ(f��)�?ji��n)对齐好看。需要注意的是首��|��者最后一��|��可能有空白页�Q�需要额外处理�?/span>

参考代码如下：(x��)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct node
{
    int front[2];
    int back[2];
}a[25];

int main(void)
{
    int n;
    int i;
    int pages;
    int t;
    while (scanf("%d", &n) != EOF && n != 0)
    {
        pages = n/4;
        if (n%4 != 0)
            ++pages;
        t = 0;
        memset(a, 0, sizeof(a));
        for (i = 0; i < pages-1; ++i)
        {

            ++t;
            a[i].front[1] = t;
            ++t;
            a[i].back[0] = t;
        }
        if (n > t)
        {
            ++t;
            a[i].front[1] = t;
        }
        if (n > t)
        {
            ++t;
            a[i].back[0] = t;
        }
        i = pages - 1;
        while (n > t+1)
        {
            ++t;
            a[i].back[1] = t;
            ++t;
            a[i].front[0] = t;
            --i;
        }
        if (n > t)
        {
            ++t;
            a[i].back[1] = t;
        }
        if (n > t)
        {
            ++t;
            a[i].front[0] = t;
        }
        printf("Printing order for %d pages:\n", n);
        for (i = 0; i < pages-1; ++i)
        {
            printf("Sheet %d, front: ", i+1);
            if (a[i].front[0] == 0)
            {
                printf("%s, ", "Blank");
            }
            else
            {
                printf("%d, ", a[i].front[0]);
            }
            if (a[i].front[1] == 0)
            {
                printf("%s\n", "Blank");
            }
            else
            {
                printf("%d\n", a[i].front[1]);
            }
            printf("Sheet %d, back : ", i+1);
            if (a[i].back[0] == 0)
            {
                printf("%s, ", "Blank");
            }
            else
            {
                printf("%d, ", a[i].back[0]);
            }
            if (a[i].back[1] == 0)
            {
                printf("%s\n", "Blank");
            }
            else
            {
                printf("%d\n", a[i].back[1]);
            }
        }
        if (a[i].front[0] != 0 || a[i].front[1] != 0)
        {
            printf("Sheet %d, front: ", i+1);
            if (a[i].front[0] == 0)
            {
                printf("%s, ", "Blank");
            }
            else
            {
                printf("%d, ", a[i].front[0]);
            }
            if (a[i].front[1] == 0)
            {
                printf("%s\n", "Blank");
            }
            else
            {
                printf("%d\n", a[i].front[1]);
            }
        }
        if (a[i].back[0] != 0 || a[i].back[1] != 0)
        {
            printf("Sheet %d, back : ", i+1);
            if (a[i].back[0] == 0)
            {
                printf("%s, ", "Blank");
            }
            else
            {
                printf("%d, ", a[i].back[0]);
            }
            if (a[i].back[1] == 0)
            {
                printf("%s\n", "Blank");
            }
            else
            {
                printf("%d\n", a[i].back[1]);
            }
        }
    }
    return 0;
}

李东�?/a> 2010-10-18 17:05 发表评论

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ZOJ_1178_Booklet Printing