C++博客-focus and persistent-文章分类-OJ

Problem 1007 - Feeding Animals(I)

kalvinyZhang — Wed, 21 Sep 2011 16:52:00 GMT

贪心算法，我不知道呀，直接就上了，先这样刷着吧，以后可能就会熟悉。要学习的点是实现多维数组的行和列分别排序

#include <stdio.h>
#define MAX 10000

int main()
{
    int n, i, j, min, sum = 0;
    int s[8][MAX];
    int t[MAX];

    scanf("%d", &n);

    for (i = 0; i <= 7; i++) {
        for (j = 0; j < n; j++)
            scanf("%d", &s[i][j]);
    }
    min = s[0][0];
    for (j = 0; j < n; j++) {
        for (i = 0; i < 8; i++) {        //寻找二维数组每列的最小值
            if (s[i][j] <= min) {
                min = s[i][j];
                t[j] = min;
            }
        }
        min = s[0][j+1]; //每一列定位的最小值，就是每列的第一个元素
    }

    for (i = 0; i < n; i++) {
        sum = sum + t[i];
    }
    printf("%d\n", sum);
    return 0;
}

kalvinyZhang 2011-09-22 00:52 发表评论

Problem 1020 - Adjacent Difference

kalvinyZhang — Wed, 21 Sep 2011 16:34:00 GMT

这个题，各种排序练习不解释呀，很好的练习基础排序的题目，现在用了插入和shell排序，可以继续快点呀，先贴着吧#include <stdio.h>

#define MAX 1000

int main()
{
    int n, m, i, j, tmp;
    int s[MAX], t[MAX];
    scanf("%d", &n);
    while (n--) {
        scanf("%d", &m);
        for (i = 0; i < m; i++)
            scanf("%d", &s[i]);
        for (i = m-1; i >= 1; i--) {
            t[i] = s[i] - s[i-1];
        }
        t[0] = s[0];
        /**
        insert sort
        for (i = 1; i < m; i++) {
            tmp = t[i];
            for (j = i; j > 0 && t[j-1] > tmp; j--)
                t[j] = t[j-1];
            t[j] = tmp;
        }
        **/
        //shellsort
        int gap;
        for (gap = m / 2; gap > 0; gap /= 2) {
            for (i = gap; i < m; i++) {
                tmp = t[i];
                for (j = i; j >= gap; j -= gap)
                    if (tmp < t[j - gap])
                        t[j] = t[j - gap];
                    else
                        break;
                t[j] = tmp;
            }
        }
        for (i = 0; i < m; i++)
            printf("%d ", t[i]);
        printf("\n");
    }
    return 0;
}

kalvinyZhang 2011-09-22 00:34 发表评论

Problem 1064 - What Is Left

kalvinyZhang — Wed, 21 Sep 2011 16:26:00 GMT

多说一句，虽然是水题，但是有个地方，真的学到了好多。就是从一个已经排序的数组中删除重复的数字。这里的算法比较烂，用的插入排序。这里输入的数据范围很大，要用到高效的排序方法。
这个题目的意思是，先输入要输入的数字，然后排序输出，其中输出的数组中不带重复的数字。下面是代码，算法没有优化。还有代码风格的问题，多用缩进和括号

#include <stdio.h>
#define MAX 1000

int main()
{
    int n, i, j, tmp;
    int s[MAX];
    int t[MAX];
    while (scanf("%d", &n) != EOF) {
        for (i = 0; i < n; i++)
            scanf("%d", &s[i]);
        for (i = 0; i < n; i++) {
            tmp = s[i];
            for (j = i; j > 0 && tmp < s[j-1]; j--)
                s[j] = s[j-1];
            s[j] = tmp;
        }
        int k = 0;
        t[k++] = s[0]; 这个地方是重点，开始k没有做加加运算，漏掉了开头或者最后的一个元素
        for (i = 1; i < n; i++) {
            if (s[i] != s[i-1])
                t[k++] = s[i];

        }
        for (i = 0; i < k; i++)
            printf("%d ", t[i]);
        printf("\n");
    }
    return 0;
}

kalvinyZhang 2011-09-22 00:26 发表评论

401 - Palindromes

kalvinyZhang — Sun, 04 Sep 2011 15:39:00 GMT

说实话有点郁闷，我不知道哪里出了问题到现在没有AC，但是基本的思想就放在那里

Palindromes

A regular palindrome is a string of numbers or letters that is the same forward as backward. For example, the string "ABCDEDCBA" is a palindrome because it is the same when the string is read from left to right as when the string is read from right to left.

A mirrored string is a string for which when each of the elements of the string is changed to its reverse (if it has a reverse) and the string is read backwards the result is the same as the original string. For example, the string "3AIAE" is a mirrored string because "A" and "I" are their own reverses, and "3" and "E" are each others' reverses.

A mirrored palindrome is a string that meets the criteria of a regular palindrome and the criteria of a mirrored string. The string "ATOYOTA" is a mirrored palindrome because if the string is read backwards, the string is the same as the original and because if each of the characters is replaced by its reverse and the result is read backwards, the result is the same as the original string. Of course, "A", "T", "O", and "Y" are all their own reverses.

A list of all valid characters and their reverses is as follows.

Character	Reverse	Character	Reverse	Character	Reverse
`A`	`A`	`M`	`M`	`Y`	`Y`
`B`		`N`		`Z`	`5`
`C`		`O`	`O`	`1`	`1`
`D`		`P`		`2`	`S`
`E`	`3`	`Q`		`3`	`E`
`F`		`R`		`4`
`G`		`S`	`2`	`5`	`Z`
`H`	`H`	`T`	`T`	`6`
`I`	`I`	`U`	`U`	`7`
`J`	`L`	`V`	`V`	`8`	`8`
`K`		`W`	`W`	`9`
`L`	`J`	`X`	`X`

Note that O (zero) and 0 (the letter) are considered the same character and therefore ONLY the letter "0" is a valid character.

Input

Input consists of strings (one per line) each of which will consist of one to twenty valid characters. There will be no invalid characters in any of the strings. Your program should read to the end of file.

Output

For each input string, you should print the string starting in column 1 immediately followed by exactly one of the following strings.

`STRING`	`CRITERIA`
" `-- is not a palindrome.`"	if the string is not a palindrome and is not a mirrored string
" `-- is a regular palindrome.`"	if the string is a palindrome and is not a mirrored string
" `-- is a mirrored string.`"	if the string is not a palindrome and is a mirrored string
" `-- is a mirrored palindrome.`"	if the string is a palindrome and is a mirrored string

Note that the output line is to include the -'s and spacing exactly as shown in the table above and demonstrated in the Sample Output below.

In addition, after each output line, you must print an empty line.

Sample Input

NOTAPALINDROME  ISAPALINILAPASI  2A3MEAS  ATOYOTA

Sample Output

NOTAPALINDROME -- is not a palindrome.   ISAPALINILAPASI -- is a regular palindrome.   2A3MEAS -- is a mirrored string.   ATOYOTA -- is a mirrored palindrome.

#include <stdio.h>

palindrome的意思就是正读反过来读都是一样的，mirrored的情况就列在上面的表上。下面是代码

#include <string.h>

const char one[] = "AEHIJLMOSTUVWXYZ12357";
const char two[] = "A3HILJMO2TUVWXY51SEZ8";
const long len = 35;

bool Palindrome(const char *s)
{
    long begin = 0, end = strlen(s)-1;
    while (begin <= end) {
        if (s[begin] == s[end]) {
            begin++;
            end--;
        }
        else
            return false;
    }
    return true;
}
long pos(char ch)
{
    for (long i = 0; i < len; i++)
        if (one[i] == ch)
            return i;
    return len;
}
bool Mirror(const char *s)
{
    long begin = 0, end = strlen(s)-1;
    while (begin <= end) {
        long tmp = pos(s[begin]);
        if (s[end] == two[tmp]) {
            begin++;
            end--;
        }
        else
            return false;
    }
    return true;
}
int main()
{
    char str[1000];
    while (scanf("%s", str) != EOF) {
        bool a = false, b = false;
        a = Palindrome(str);
        b = Mirror(str);
        if (a && b)
            printf("%s -- is a mirrored palindrome.\n", str);
        else if (a)
            printf("%s -- is a regular palindrome.\n", str);
        else if (b)
            printf("%s -- is a mirrored string.\n", str);
        else
            printf("%s -- is not a palindrome.\n", str);
        putchar('\n');
    }
    return 0;
}
是用了CPP写的代码，应该是C风格的CPP，因为我是先看了部分CPP最后不爽转到了C嘛。。。基本的思想是这样的，首先，第一个函数Palindrome，两个参数，分别正反读取字符串，这个思想以后可以常用。然后Mirroed和pos函数之间的关系是这样的，pos函数主要是确定字符串中的字符在列表中的位置，然后拿去Mirroed中和第二个数组比较，最终输出结果，我不知道这个程序哪里出了问题ing，慢慢学习，也许几个星期之后。。。我能用纯C直接写出来了。。。

kalvinyZhang 2011-09-04 23:39 发表评论

1071 辉辉的一天

kalvinyZhang — Fri, 02 Sep 2011 17:08:00 GMT

水题，但是有一个地方导致了不能AC，多个不等的时候if (!(a == 0 && b == 0 && c == 0 && d == 0 && e == 0 && f == 0)) -这个写法是正确的第一次写成了
if ((a != 0&& b != 0 && c != 0 && d != 0 && e != 0 && f != 0))这个写法是没有办法AC的，至于为什么应该进一步深入应该会慢慢意识到。

1 #include <stdio.h>
2
3 int main()
4 {
5     int a, b, c, d, e, f;
6     while (scanf("%d%d%d%d%d%d", &a, &b, &c, &d, &e, &f)) {
7         if (!(a == 0 && b == 0 && c == 0 && d == 0 && e == 0 && f == 0)) {
8             double sum;
9             sum = 3600*(d+12-a) + (e-b)*60 + f-c;
10             printf("%.0lf\n", sum);
11         }
12         else
13             break;
14     }
15     return 0;
16 }

代码就放在这里吧。。。其中的double 类型可以换成long int，但是double似乎更适合一些

kalvinyZhang 2011-09-03 01:08 发表评论