hdu 3746 Cyclic Nacklace
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3746
题目大意:补充珠子数使其成为手链,手链的规格是:比如这一组数据:abca,要想成为手链,必须满足abcabc,还要加两个,所以输出2。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 char str[1000005]; 6 int next[100005],len; 7 8 void get_next() 9 {10 int i=0,j=-1;11 next[0]=-1;12 len=strlen(str);13 while (i<len)14 {15 if (j==-1||str[i]==str[j])16 {17 i++;18 j++;19 next[i]=j;20 }21 else22 j=next[j];23 }24 }25 26 int main ()27 {28 int t;29 scanf ("%d",&t);30 while (t--)31 {32 scanf("%s",str);33 get_next();34 //for (int i=0;i<len;i++)35 //cout<<next[i]<<endl; 36 int q=len-next[len];37 if (len%q==0&&len!=q)38 printf ("0\n");39 else40 printf("%d\n",q-len%q);41 }42 return 0;43 }
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