| 摘要 |
PROBLEM TO BE SOLVED: To efficiently make a jump to a point of time that a user desires. SOLUTION: The user does a 1st skip operation two seconds (time lag) after a CM starts. Consequently, a skip is made by a skip time of 27 (sec.). Then a 2nd skip operation is done within a specified time. Consequently, a skip is made by a skip time of 30 (sec.). Similarly, when a 3rd skip operation is done within the specified time, a skip is made by a skip time of 30 (sec.). Similarly, when a 4th skip operation is done within the specified time, a skip is made by a skip time of 30 (sec.). Through those successive skip operations, a jump to a point of time that the user desires (1 (sec.) before the program restarts) can be made. COPYRIGHT: (C)2005,JPO&NCIPI
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