Pro.ID1375 TitlePoor contestant Prob Title链接http://10.20.2.8/oj/exercise/problem?problem_id=1375 AC5 Submit14 Ratio35.71% 时间&空间限制描述As everybody known, "BG meeting" is very very popular in the ACM training team of ZSU. After each online contest, they will go out for "smoking". Who will be the poor ones that have to BG the others? Of course, the half who solve less problems. The rule runs well when the number of the contestants is even. But if the number is odd, it is impossible to divide them into two equal parts. It gives a dilemma to the BG meeting committee. After a careful discussion with Mr. Guo, a new rule emerged: if the number of the contestant is odd, the committee will first sort the contestants according to the number of problems they solved, and then they will pick out the middle one. This poor boy or girl will have no chance to attend the BG meeting. Strange rule, isn't it? As the number of the contestants is becoming more and more large, the committee need to write a program which will pick out the poor one efficiently. Note that: Every contestant solves different number of problems. The total number of the contestants will not exceed 105. 输入There are several cases in the input. The first line of the input will be an integer M, the number of the cases. Each case is consisted of a list of commands. There are 3 types of commands. 1. Add xxx n : add a record to the data base, where xxx is the name of the contestant, which is only consisted of at most 10 letters or digits, n is the number of problems he/she solved. (Each name will appear in Add commands only once). 2. Query : 3. End :End of the case. 输出Description As everybody known, "BG meeting" is very very popular in the ACM training team of ZSU. After each online contest, they will go out for "smoking". Who will be the poor ones that have to BG the others? Of course, the half who solve less problems. The rule runs well when the number of the contestants is even. But if the number is odd, it is impossible to divide them into two equal parts. It gives a dilemma to the BG meeting committee. After a careful discussion with Mr. Guo, a new rule emerged: if the number of the contestant is odd, the committee will first sort the contestants according to the number of problems they solved, and then they will pick out the middle one. This poor boy or girl will have no chance to attend the BG meeting. Strange rule, isn't it? As the number of the contestants is becoming more and more large, the committee need to write a program which will pick out the poor one efficiently. Note that: Every contestant solves different number of problems. The total number of the contestants will not exceed 105. Input There are several cases in the input. The first line of the input will be an integer M, the number of the cases. Each case is consisted of a list of commands. There are 3 types of commands. 1. Add xxx n : add a record to the data base, where xxx is the name of the contestant, which is only consisted of at most 10 letters or digits, n is the number of problems he/she solved. (Each name will appear in Add commands only once). 2. Query : 3. End :End of the case. Output 1. For the Query command: If the current number of contestants is odd, the program should output the poor contestant's name currently even if there is only one contestants, otherwise, just out put "No one!" (without quotes). 2. For the End command: If the total number of contestants in the data base is even, you should out put "Happy BG meeting!!" (without quotes), otherwise, you should out put the "xxx is so poor." (without quotes) where xxx is the name of the poor one. 3. Each case should be separated by a blank line. Sample Input 2 Sample Output No one! 样例输入2 样例输出No one! 提示作者 |