1.1 Broken Necklace

是我遇到的比较复杂的模拟了0 0  

这道题先是自己写的糟糕版,交上去WA了,跟着数据改..最后改得面目全非....昨天找标程交的!今天稍微修改了一下..就过了T T

这题对我来说没考虑到的是以下两点:

1.在枚举每个端口,统计最大长度的时候,没有想到要是全是某个字母,那不是不停转圈.

(另:我在编写此功能函数时,如果判重也很无力,等等看看标程)

2.就算考虑上面那一点,遇到全是某个字母的情况,会重复统计,我在最后判断统计长度大于输入长度则输出输入的长度

 标程重要的部分:

1.定义更为完整的mod,而我是每次mod各种判断,麻烦.

2.统计函数很值得我借鉴,首先,用dir储存方向,要单独处理的只是初始位置,接着是判重,有点考逻辑吧 = =

 

/* 
 * Return n mod m.  The C % operator is not enough because
 * its behavior is undefined on negative numbers.
 */
int 
mod(int n, int m)
{
    while(n < 0)
    n += m;
    return n%m;
}

/*
 * Calculate number of beads gotten by breaking
 * before character p and going in direction dir,
 * which is 1 for forward and -1 for backward.
 */
int
nbreak(int p, int dir)
{
    char color;
    int i, n;

    color = 'w';

    /* Start at p if going forward, bead before if going backward */
    if(dir > 0)
    i = p;
    else
    i = mod(p-1, len);

    /* We use "n<len" to cut off loops that go around the whole necklace */
    for(n=0; n<len; n++, i=mod(i+dir, len)) {
    /* record which color we're going to collect */
    if(color == 'w' && necklace[i] != 'w')
        color = necklace[i];

    /* 
     * If we've chosen a color and see a bead
     * not white and not that color, stop 
     */
    if(color != 'w' && necklace[i] != 'w' && necklace[i] != color)
        break;
    }
    return n;
}

 

/*
ID:y7276571
LANG: C
TASK: beads
*/
#include<stdio.h>
#include<string.h>
#define MAXLEN 350
int n,left[MAXLEN] = {0}, right[MAXLEN] = {0};
char beads[MAXLEN];
int getpos(int pos)
{
    if(pos < 0) pos = n+pos;
    else if(pos >= n) pos -= n;
    return pos;
}
int maxlen_left(int pos)
{
    int ans = 1, stop = pos;
    char c;
    while(stop - pos <= n)
    {
        if(beads[getpos(--pos)] == 'w') ans++;
        else {c = beads[getpos(pos)]; break;}
    }
    while(stop - pos <= n)
    {
        if(beads[getpos(--pos)] == 'w') { ans++; continue; }
        else
        {
            if(beads[getpos(pos)] == c) ans++;
            else return ans;
        }
    }
}
int maxlen_right(int pos)
{
    int ans = 1, stop = pos;
    char c;
    while(pos - stop < n)
    {
        if(beads[getpos(pos++)] == 'w') ans++;
        else {c = beads[getpos(pos-1)]; break;}
    }
    while(pos - stop < n)
    {
        if(beads[getpos(pos++)] == 'w') { ans++; continue; }
        else
        {
            if(beads[getpos(pos-1)] == c) ans++;
            else return ans;
        }
    }
}
int main(void)
{
    freopen("beads.in", "r", stdin);
    freopen("beads.out", "w", stdout);
    int i, j = 0, maxlen = 0;
    scanf("%d%s", &n, beads);
    for(i = 0; i < n; i++)
    {
        left[i] = maxlen_left(i);
        right[i] = maxlen_right(i);
    }
    for(i = 0; i < n && !j; i++)    maxlen = maxlen < left[i]+right[i] ?left[i]+right[i] : maxlen;
    if(maxlen > n) printf("%d\n", n);
    else    printf("%d\n", maxlen);
    exit(0);
}