十分详细的解释：
几道相关例题的代码：
1 #include2 #include 3 #include 4 #include 5 6 using namespace std; 7 8 char buf[100001]; 9 10 int minExp(char *s) {11 int i = 0, j = 1, k = 0, t;12 int len = strlen(s);13 14 while (i < len && j < len && k < len) {15 t = s[(i + k) % len] - s[(j + k) % len];16 if (!t) {17 k++;18 } else {19 if (t > 0) i = i + k + 1;20 else {21 j = j + k + 1;22 }23 if (i == j) j++;24 k = 0;25 }26 //printf("%d %d %d\n", i, j, k);27 }28 29 return min(i, j);30 }31 32 int main() {33 int T, n;34 35 scanf("%d", &T);36 while (T-- && ~scanf("%d %s", &n, buf)) {37 printf("%d\n", minExp(buf));38 memset(buf, 0, sizeof(buf));39 }40 41 return 0;42 }
1 #include2 #include 3 #include 4 #include 5 6 using namespace std; 7 8 char buf[10001]; 9 10 int minExp(char *s) {11 int i = 0, j = 1, k = 0, t;12 int len = strlen(s);13 14 while (i < len && j < len && k < len) {15 t = s[(i + k) % len] - s[(j + k) % len];16 if (!t) {17 k++;18 } else {19 if (t > 0) i = i + k + 1;20 else {21 j = j + k + 1;22 }23 if (i == j) j++;24 k = 0;25 }26 //printf("%d %d %d\n", i, j, k);27 }28 29 return min(i, j) + 1;30 }31 32 int main() {33 int T;34 35 scanf("%d", &T);36 while (T-- && ~scanf("%s", buf)) {37 printf("%d\n", minExp(buf));38 memset(buf, 0, sizeof(buf));39 }40 41 return 0;42 }
1 #include2 #include 3 #include 4 #include 5 #include 6 #include 7 8 using namespace std; 9 10 set cnt;11 char buf[101];12 13 int minExp(char *s) {14 int i = 0, j = 1, k = 0, t;15 int len = strlen(s);16 17 while (i < len && j < len && k < len) {18 t = s[(i + k) % len] - s[(j + k) % len];19 if (!t) {20 k++;21 } else {22 if (t > 0) i = i + k + 1;23 else {24 j = j + k + 1;25 }26 if (i == j) j++;27 k = 0;28 }29 //printf("%d %d %d\n", i, j, k);30 }31 32 return min(i, j);33 }34 35 void con(char *a){36 int pos = minExp(a);37 int len = strlen(a);38 char tmp[101];39 40 strcpy(tmp, a);41 for (int i = 0; i < len; i++){42 a[i] = tmp[(pos + i) % len];43 }44 }45 46 int deal(int n){47 cnt.clear();48 for (int i = 0; i < n; i++){49 scanf("%s", buf);50 con(buf);51 cnt.insert(buf);52 }53 54 return cnt.size();55 }56 57 int main(){58 int n;59 60 while (~scanf("%d", &n)){61 printf("%d\n", deal(n));62 }63 64 return 0;65 }
1 #include2 #include 3 #include 4 #include 5 #include 6 #include 7 8 using namespace std; 9 10 set cnt;11 char buf[300001], tmp[300001];12 13 int minExp(char *s) {14 int i = 0, j = 1, k = 0, t;15 int len = strlen(s);16 17 while (i < len && j < len && k < len) {18 t = s[(i + k) % len] - s[(j + k) % len];19 if (!t) {20 k++;21 } else {22 if (t > 0) i = i + k + 1;23 else {24 j = j + k + 1;25 }26 if (i == j) j++;27 k = 0;28 }29 //printf("%d %d %d\n", i, j, k);30 }31 32 return min(i, j);33 }34 35 void work() {36 int len = strlen(buf);37 38 tmp[0] = (buf[0] - buf[len - 1] + 8) % 8 + '0';39 for (int i = 1; i < len; i++){40 tmp[i] = (buf[i] - buf[i - 1] + 8) % 8 + '0';41 }42 43 int pos = minExp(tmp);44 45 for (int i = 0; i < len; i++) {46 buf[i] = tmp[(pos + i) % len];47 }48 }49 50 int main() {51 while (~scanf("%s", buf)) {52 work();53 printf("%s\n", buf);54 }55 56 return 0;57 }
——written by Lyon