杭电5499 SDOI
来源:互联网 发布:阿里云大数据平台 编辑:程序博客网 时间:2024/05/11 18:02
SDOI
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 691 Accepted Submission(s): 271
Problem Description
The Annual National Olympic of Information(NOI) will be held.The province of Shandong hold a Select(which we call SDOI for short) to choose some people to go to the NOI. n(n≤100) people comes to the Select and there is m(m≤50) people who can go to the NOI.
According to the tradition and regulation.There were two rounds of the SDOI, they are so called "Round 1" and "Round 2", the full marks of each round is300 .
All the n people take part in Round1 and Round2, now the original mark of every person is known. The rule of SDOI of ranking gets to the "standard mark". For each round there is a highest original mark,let's assume that isx .(it is promised that not all person in one round is 0,in another way,x>0 ). So for this round,everyone's final mark equals to his/her original mark∗(300/x) .
After we got everyone's final mark in both round.We calculate the Ultimate mark of everyone as0.3∗round1′s final mark + 0.7∗round2′s final mark.It is so great that there were no two persons who have the same Ultimate mark.
After we got everyone's Ultimate mark.We choose the persons as followed:
To encourage girls to take part in the Olympic of Information.In each province,there has to be a girl in its teams.
1. If there is no girls take part in SDOI,The boys with the rank of first m enter the team.
2. If there is girls, then the girl who had the highest score(compared with other girls) enter the team,and other(boys and other girls) m-1 people with the highest mark enter the team.
Just now all the examination had been finished.Please write a program, according to the input information of every people(Name, Sex ,The original mark of Round1 and Round2),Output the List of who can enter the team with their Ultimate mark decreasing.
According to the tradition and regulation.There were two rounds of the SDOI, they are so called "Round 1" and "Round 2", the full marks of each round is
All the n people take part in Round1 and Round2, now the original mark of every person is known. The rule of SDOI of ranking gets to the "standard mark". For each round there is a highest original mark,let's assume that is
After we got everyone's final mark in both round.We calculate the Ultimate mark of everyone as
After we got everyone's Ultimate mark.We choose the persons as followed:
To encourage girls to take part in the Olympic of Information.In each province,there has to be a girl in its teams.
1. If there is no girls take part in SDOI,The boys with the rank of first m enter the team.
2. If there is girls, then the girl who had the highest score(compared with other girls) enter the team,and other(boys and other girls) m-1 people with the highest mark enter the team.
Just now all the examination had been finished.Please write a program, according to the input information of every people(Name, Sex ,The original mark of Round1 and Round2),Output the List of who can enter the team with their Ultimate mark decreasing.
Input
There is an integer T(T≤100) in the first line for the number of testcases and followed T testcases.
For each testcase, there are two integersn and m in the first line(n≥m) , standing for the number of people take part in SDOI and the allowance of the team.Followed with n lines,each line is an information of a person. Name(A string with length less than 20 ,only contain numbers and English letters),Sex(male or female),the Original mark of Round1 and Round2 (both equal to or less than 300 ) separated with a space.
For each testcase, there are two integers
Output
For each testcase, output "The member list of Shandong team is as follows:" without Quotation marks.
Followedm lines,every line is the name of the team with their Ultimate mark decreasing.
Followed
Sample Input
210 8dxy male 230 225davidwang male 218 235evensgn male 150 175tpkuangmo female 34 21guncuye male 5 15faebdc male 245 250lavender female 220 216qmqmqm male 250 245davidlee male 240 160dxymeizi female 205 1902 1dxy male 300 300dxymeizi female 0 0
Sample Output
The member list of Shandong team is as follows:faebdcqmqmqmdavidwangdxylavenderdxymeizidavidleeevensgnThe member list of Shandong team is as follows:dxymeiziHintFor the first testcase: the highest mark of Round1 if 250,so every one's mark times(300/250)=1.2, it's same to Round2.The Final of The Ultimate score is as followedfaebdc 298.20qmqmqm 295.80davidwang 275.88dxy 271.80lavender 260.64dxymeizi 233.40davidlee 220.80evensgn 201.00tpkuangmo 29.88guncuye 14.40For the second testcase,There is a girl and the girl with the highest mark dxymeizi enter the team, dxy who with the highest mark,poorly,can not enter the team.
Source
BestCoder Round #59 (div.2)
Recommend
hujie | We have carefully selected several similar problems for you: 5589 5588 5587 5586 5585
结构体水题,这两天心情稍微 有点小缓和 ,做两道水题:
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct node{char name[1100];int sex;double mark1;//round 1double mark2;//round 2double mark3;//scoor;}t[110000];int cmp(node a,node b){return a.mark3 > b.mark3;}int i,j,p,m,n,k,l,help;char s[110];int main(){scanf("%d",&p);while(p--){scanf("%d%d",&m,&n);double Max1=0;double Max2=0;for(i=0;i<m;i++){scanf("%s%s%lf%lf",t[i].name,s,&t[i].mark1,&t[i].mark2);if(s[0]=='f')t[i].sex=0;//gorlelset[i].sex=1;//boyMax1=max(Max1,t[i].mark1);Max2=max(Max2,t[i].mark2);}for(i=0;i<m;i++)t[i].mark3=t[i].mark1*300/Max1*0.3+t[i].mark2*300/Max2*0.7;sort(t,t+m,cmp);int flag=0;help=-1;for(i=0;i<n;i++)if(t[i].sex==0)flag=1;for(i=0;i<m;i++)if(t[i].sex==0){help=i;break;}printf("The member list of Shandong team is as follows:\n");////for(i=0;i<m;i++)//printf("+++%lf\n",t[i].mark3)if(flag||help==-1)//木有女生或者女生已经选上 for(i=0;i<n;i++)printf("%s\n",t[i].name);else//有女生但是没选上 {for(i=0;i<n-1;i++)printf("%s\n",t[i].name);printf("%s\n",t[help].name);}}}
0 0
- 杭电5499 SDOI
- 【杭电】[5499]SDOI
- 【杭电 5499】SDOI
- 杭电-5499SDOI(结构体排序)
- 【杭电oj】5499 - SDOI(结构体排序,水)
- 【杭电-oj】-5499-SDOI(结构体)
- 5499 SDOI
- HDU 5499 SDOI
- HDU 5499 SDOI
- hdoj 5499 SDOI 【拼手速】
- hdu 5499 SDOI(水)
- HDOJ 5499 SDOI (乱搞)
- HDOJ 5499 SDOI
- HDU 5499:SDOI【排序】
- hdoj 5499 SDOI
- HDU 5499 SDOI
- HDUoj 5499 SDOI (贪心
- SDOI
- git之忽略文件
- HDU 1698 Just a Hook(线段树的区间更新《标记》)
- Android 开发必备知识:我和 Gradle 有个约会
- ListView工作原理及异步加载图片乱序问题
- 关于Ubuntu下ibus在firefox浏览器中选中即删除的解决办法
- 杭电5499 SDOI
- iOS 国际化
- 差分约束学习(二)POJ 1275:Cashier Employment
- 查找——图文详解HashTree(哈希树)
- session cursor
- window对象
- 关于liunx上面的vi编辑器的使用小总结
- 0707什么是JSON+如何处理JSON字符串0
- 扣丁学堂——Activity(一)