導入Java . io . buffered reader；
導入Java . io . io exception；
導入Java . io . inputstreamreader；
公開課問題壹{
/**
*壹組序列由AGCT的四個字符組成，如:taacgagagagataaaaaaaaaaaaaaatgaggaaaaaaaaaaaaaaaagatggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaaggaag
*，請求壹個小程序。
*例如，輸入壹個字符串AAAG，並從頭開始。如果這個字符串是上面的序列，輸出:成功找到，記錄位置。如果沒有上述字符串，輸出:找不到，並退出。
*
* @param args
*/
公共靜態void main(String[] args) {
string data = " taacgagagagagattaaaaacaagaatgattgaaggacagaaaaggaaaaaaaagaagaacaatgatggataaggaggaggtggatgacaagattaaggaggaggagagataatagaagaaggattgaaggaagggaggaaggaaggaaaaaaaaaaaaaagaggaagaggagagat "；
System.out.println("請輸入壹個字符串，回車結束")；
buffered reader reader = new buffered reader(new InputStreamReader(
system . in))；
String line = null
嘗試{
line = reader . readline()；
} catch (IOException e) {
e . printstacktrace()；
}
int pos = data . index of(line)；
if (pos == -1) {
System.out.println("找不到出口")；
}否則{
System.out.println ("found，in first "+(pos+1)+"字符")；
}
}
}