UVa 10340 - All in All
2015-08-16 23:19
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https://uva.onlinejudge.org/external/103/10340.pdf
10340 All in All
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
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
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
分析:
给两个串,让你判断能否在第二个串中(不改变顺序)去掉一些字符(也可以不去),得到第一个串。
直接匹配。
复杂度:O(len2)
10340 All in All
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
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
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
分析:
给两个串,让你判断能否在第二个串中(不改变顺序)去掉一些字符(也可以不去),得到第一个串。
直接匹配。
复杂度:O(len2)
#include <iostream> #include <sstream> #include <iomanip> #include <vector> #include <deque> #include <list> #include <set> #include <map> #include <stack> #include <queue> #include <bitset> #include <string> #include <numeric> #include <algorithm> #include <functional> #include <iterator> #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <cctype> #include <complex> #include <ctime> #define INF 0x3f3f3f3f #define eps 1e-6 typedef long long LL; const double pi = acos(-1.0); const long long mod = 1e9 + 7; using namespace std; char s1[100005]; char s2[100005]; int main() { //freopen("int.txt","r",stdin); //freopen("out.txt","w",stdout); while(scanf("%s %s",s1,s2) == 2) { int len1 = strlen(s1); int len2 = strlen(s2); int i = 0; //printf("%s\n%s\n",s1,s2); for(int j = 0;j < len2;j++) { if(s1[i] == s2[j]) i++; if(i == len1) break; } if(i == len1) puts("Yes"); else puts("No"); } return 0; }
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