好久不交题了,而且发现最近状态恶劣的不行,AC率恶心到不行 ~
这个题代码有点长,但实际思路很简单,qsort 拍一下序,然后,将最长的字符串与最短的字符串搭配,因为一个字符串最多分成两个不同的最长的字符串,所以最长与最短子字符串搭配最多8种情况,枚举出来,其中有一个就是所求的答案。
代码如下:
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;
int cmp(const void *a, const void *b)
{
int size = strlen((char*)a) - strlen((char*)b);
if(!size)
return strcmp((char*)a, (char*)b);
return size;
}
int main()
{
#ifdef st
freopen("sample.txt", "r", stdin);
#endif
char a[150][260], b[260], c[260];
int num;
scanf("%d",&num);
getchar();
getchar();
while(num--)
{
int ct = 0;
memset(a, 0, sizeof(a));
while(1)
{
gets(a[ct]);
if(!strlen(a[ct]))
break;
++ct;
}
qsort(a, ct, sizeof(a[0]), cmp);
int flag = 0;
if(strlen(a[0]) == strlen(a[1]))
{
sprintf(b,"%s%s",a[0],a[ct - 1]);
sprintf(c,"%s%s",a[ct - 2],a[1]);
if(!strcmp(b, c)) // 两个字符串相同说明是所求的完整字符串
flag = 1;
if(!flag)
{
sprintf(b,"%s%s",a[1],a[ct - 1]);
sprintf(c,"%s%s",a[ct - 2],a[0]);
if(!strcmp(b, c))
flag = 1;
}
if(!flag)
{
sprintf(b,"%s%s",a[ct - 1],a[0]);
sprintf(c,"%s%s",a[1],a[ct - 2]);
if(!strcmp(b, c))
flag = 1;
}
if(!flag)
{
sprintf(b,"%s%s",a[ct - 1],a[1]);
sprintf(c,"%s%s",a[0],a[ct - 2]);
if(!strcmp(b, c))
flag = 1;
}
if(!flag)
{
sprintf(b,"%s%s",a[0],a[ct - 1]);
sprintf(c,"%s%s",a[1],a[ct - 2]);
if(!strcmp(b, c))
flag = 1;
}
if(!flag)
{
sprintf(b,"%s%s",a[1],a[ct - 1]);
sprintf(c,"%s%s",a[0],a[ct - 2]);
if(!strcmp(b, c))
flag = 1;
}
if(!flag)
{
sprintf(b,"%s%s",a[ct - 1],a[0]);
sprintf(c,"%s%s",a[ct - 2],a[1]);
if(!strcmp(b, c))
flag = 1;
}
if(!flag)
{
sprintf(b,"%s%s",a[ct - 1],a[1]);
sprintf(c,"%s%s",a[ct - 2],a[0]);
if(!strcmp(b, c))
flag = 1;
}
}
else
sprintf(b,"%s%s",a[ct - 1],a[0]);
printf("%s\n",b);
if(num)
puts("");
}
return 0;
}