coj 1100 Magic Spell

这道题的HINT彻底把我弄糊涂了，结果WA5次才发现是求最大子序列而不是连续子串：

HINT

子序列的定义：字符串S中从S[ i ]...S[ j ]的长度为K的字符(i<=j)，k<=j-i+1。

# include <stdio.h>

# include <memory.h>



# define MAX(x, y) ((x)>(y) ? (x):(y))



# define MAXN 1020



int len1, len2;

char s1[MAXN], s2[MAXN];

short int f[MAXN][MAXN];



int main()

{

    int i, j, max;



    while (~scanf("%d%s%d%s", &len1, s1+1, &len2, s2+1))

    {

        max = 0;

        memset(f, 0, sizeof(f));



        for ( i = 1; i <= len1; ++i)

            for ( j = 1; j <= len2; ++j)

            {

                if (s1[i] == s2[j]) f[i][j] = f[i-1][j-1]+1;

                else f[i][j] = MAX(f[i-1][j], f[i][j-1]);

            }



        printf("%d\n", f[len1][len2]);

    }



    return 0;

}

不过也顺便复习了一下最大子串的方法，很相似， 都可以使用滚动数组。

# include <stdio.h>

# include <memory.h>

# include <ctype.h>



# define MAXN 1020



int len1, len2;

char s1[MAXN], s2[MAXN];

short int f[MAXN][MAXN];



int main()

{

    int i, j, max;=



    while (~scanf("%d%s%d%s", &len1, s1+1, &len2, s2+1))

    {

        max = 0;

        memset(f, 0, sizeof(f));



        for ( i = 1; i <= len1; ++i)

            for ( j = 1; j <= len2; ++j)

            {

                if (tolower(s1[i]) == tolower(s2[j])) f[i][j] = f[i-1][j-1]+1;

                if (max < f[i][j]) max = f[i][j];

            }



        printf("%d\n", max);

    }



    return 0;

}