I -SDOIThe 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)n(n≤100) people comes to the Select and there is m(m≤50)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 is 300300. 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 xx.(it is promised that not all person in one round is 0,in another way,x>0x>0). So for this round,everyone's final mark equals to his/her original mark?(300/x)mark?(300/x). After we got everyone's final mark in both round.We calculate the Ultimate mark of everyone as 0.3?round1′s0.3?round1′s final mark + 0.7?round2′s0.7?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. InputThere is an integer T(T≤100)T(T≤100) in the first line for the number of testcases and followed TT testcases. For each testcase, there are two integers nn and mm in the first line(n≥m)(n≥m), standing for the number of people take part in SDOI and the allowance of the team.Followed with nn lines,each line is an information of a person. Name(A string with length less than 2020,only contain numbers and English letters),Sex(male or female),the Original mark of Round1 and Round2 (both equal to or less than 300300) separated with a space. OutputFor each testcase, output "The member list of Shandong team is as follows:" without Quotation marks. Followed mm lines,every line is the name of the team with their Ultimate mark decreasing.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 0Sample 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 th#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct stu{char name[22];char sex[10];double or1;double or2;double fr1;double fr2;double f;int l;}data[100+11];bool cmp1(stu A,stu B){return A.or1>B.or1;}bool cmp2(stu A,stu B){return A.f>B.f;}bool cmp3(stu A,stu B){return A.or2>B.or2;}int main(){int T,n,m,i,j;double x1,x2;scanf("%d",&T);while(T--){j=-1;scanf("%d%d",&n,&m);for(i=0;i<n;i++){scanf("%s %s %lf %lf",data[i].name,data[i].sex,&data[i].or1,&data[i].or2);}sort(data,data+n,cmp1);x1=data[0].or1;sort(data,data+n,cmp3);x2=data[0].or2;for(i=0;i<n;i++){data[i].fr1=data[i].or1*(300/x1);data[i].fr2=data[i].or2*(300/x2);data[i].f=data[i].fr1*0.3+data[i].fr2*0.7;data[i].l=strlen(data[i].sex);}sort(data,data+n,cmp2);for(i=0;i<n;i++){if(data[i].l==6){j=i;break;}}printf("The member list of Shandong team is as follows:/n");if(j==-1||j>=0&&j<m){for(i=0;i<m;i++){printf("%s/n",data[i].name);}}else if(j>=m){for(i=0;i<m-1;i++){printf("%s/n",data[i].name);}printf("%s/n",data[j].name);}}return 0;}