UVA 10340 - All in All

Problem E

All in All

Input: standard input

Output: standard output

Time Limit: 2 seconds

Memory Limit: 32 MB

You have devised a new encryption technique which encodes a message by inserting between its characters randomly generated strings in a clever way. Because of pending patent issues we will not discuss in detail how the strings are generated and inserted into the original message. To validate your method, however, it is necessary to write a program that checks if the message is really encoded in the final string.

Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can remove characters from t such that the concatenation of the remaining characters is s.

Input Specification

The input contains several testcases. Each is specified by two strings s, t of alphanumeric ASCII characters separated by whitespace. Input is terminated by EOF.

Output Specification

For each test case output, if s is a subsequence of t.

Sample Input

sequence subsequence

person compression

VERDI vivaVittorioEmanueleReDiItalia

caseDoesMatter CaseDoesMatter

Sample Output

Yes

No

Yes

No

---

Source: ULM Local Contest

 

题意：给字符串s和t，问t能否通过删除0个或多个字符(其他顺序不变)得到s。

 

 

#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
#include <map>
#include <queue>
#include <set>

using namespace std;

#ifdef WIN
typedef __int64 LL;
#define form "%I64d\n"
#else
typedef long long LL;
#define form "%lld\n"
#endif

#define SI(a) scanf("%d", &(a))
#define SDI(a, b) scanf("%d%d", &(a), &(b))
#define S64I scanf(form, &a)
#define SS(a) scanf("%s", (a))
#define SDS(a, b) scanf("%s%s", (a), (b))
#define SC(a) scanf("%c", &(a))
#define PI(a) printf("%d\n", a)
#define PS(a) puts(a)
#define P64I(a) printf(form, a)
#define Max(a, b) (a > b ? a : b)
#define Min(a, b) (a < b ? a : b)
#define MSET(a, b) (memset(a, b, sizeof(a)))
#define Mid(L, R) (L + (R - L)/2)
#define Abs(a) (a > 0 ? a : -a)
#define REP(i, n) for(int i=0; i < (n); i++)
#define FOR(i, a, n) for(int i=(a); i <= (n); i++)
const int INF = 0x3f3f3f3f;

const int maxn = 100000 + 50;
char t[maxn], s[maxn];

int main() {

    while(SDS(s, t) == 2) {
        int lens = strlen(s);
        int lent = strlen(t);
        int i=0, j=0;

        while(1) {
            while(i < lent && t[i] != s[j]) i++;
            if(i++ == lent) break;
            if(++j == lens) break;
        }

        if(j == lens) puts("Yes");
        else puts("No");
    }

    return 0;
}