hnuun 11544 小明的烦恼——找字符串(字符串)

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <vector>
 5 #include <cstring>
 6 #include <string>
 7 #include <algorithm>
 8 #include <string>
 9 #include <set>
10 #include <functional>
11 #include <numeric>
12 #include <sstream>
13 #include <stack>
14 #include <map>
15 #include <queue>
16 #pragma comment(linker, "/STACK:102400000,102400000")
17 #define CL(arr, val)    memset(arr, val, sizeof(arr))
18 
19 #define ll long long
20 #define inf 0x7f7f7f7f
21 #define lc l,m,rt<<1
22 #define rc m + 1,r,rt<<1|1
23 #define pi acos(-1.0)
24 
25 #define L(x)    (x) << 1
26 #define R(x)    (x) << 1 | 1
27 #define MID(l, r)   (l + r) >> 1
28 #define Min(x, y)   (x) < (y) ? (x) : (y)
29 #define Max(x, y)   (x) < (y) ? (y) : (x)
30 #define E(x)        (1 << (x))
31 #define iabs(x)     (x) < 0 ? -(x) : (x)
32 #define OUT(x)  printf("%I64d\n", x)
33 #define lowbit(x)   (x)&(-x)
34 #define Read()  freopen("a.txt", "r", stdin)
35 #define Write() freopen("b.txt", "w", stdout);
36 #define maxn 1010
37 #define maxv 1010
38 #define mod 1000000000
39 using namespace std;
40 char str[5000010];
41 int work(int m)
42 {
43     int i,j,l;
44     i=0; j=1;
45     while(i<m && j<m)
46     {
47         for(l=0;l<m;l++)
48             if(str[(i+l)%m]!=str[(j+l)%m]) break;
49         if(l>m) break;
50         if(str[(i+l)%m] > str[(j+l)%m])
51             i=i+l+1;
52         else
53             j=j+l+1;
54         if(i==j) j=i+1;
55     }
56     if(i<j) return i;
57     return j;
58 }
59 
60 int main()
61 {
62     //freopen("a.txt","r",stdin);
63     int t;
64     scanf("%d",&t);
65     while(t--)
66     {
67         scanf("%s",str);
68        // printf("%s\n",str);
69         int l=strlen(str);
70         printf("%d\n",work(l));
71     }
72     return 0;
73 }

 1 int work(int len,char pat[])  //最大表示法
 2 {
 3    //int len = strlen(pat);
 4    int i=0,j=1,k=0;
 5    while(i<len && j<len && k<len)
 6    {
 7        int t = pat[(i+k)%len] - pat[(j+k)%len];
 8        if(!t) k++;
 9        else
10        {
11            if(t>0) j = j+k+1;
12            else i = i+k+1;
13            if(i == j) j++;
14            k = 0 ;
15        }
16    }
17    return i<j?i:j;
18 }