hdu5414 (2015 + schools)--CRB and string (string match)

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The main topic: there is a A, a, two string, there is now an operation can be any one of a character x after the addition of a character y (x! =y), ask if you can change a to B.

First, if a can become B, then a must be a sub-sequence of B, which can be calculated in O (n+m) time.

If a is a sub-sequence of B, it is not possible to determine whether the added characters contain an increase in the case, we only need to judge B from the beginning of a continuous sequence of the same string, is not at the beginning of a also exist, if present, then can be converted from A to B.

#include <cstdio> #include <cstring> #include <algorithm>using namespace std; char s1[100010], s2[ 100010]; int main () {    int T, I, J, L1, L2;    char ch;    Freopen ("1009.in", "R", stdin);    Freopen ("9.out", "w", stdout);    scanf ("%d", &t);    while (t--) {        scanf ("%s%s", S1, S2);        L1 = strlen (S1);        L2 = strlen (s2);        i = j = 0;        while (I < L1 && J < L2) {            if (s1[i] = = S2[j]) {                i++; j + +;            }            else                j + +;        }        if (I < L1) {            printf ("no\n");            Continue;        }        for (j = 1; j < L2; J + +)            if (s2[j]! = S2[j-1]) break;        for (i = 0; i < J; i++)            if (s1[i]! = s2[0]) break;        if (i < j)            printf ("no\n");        else            printf ("yes\n");    }    return 0;}

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hdu5414 (2015 + schools)--CRB and string (string match)