杭电acm 5499 和 5500

SDOITime Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 76 Accepted Submission(s): 35Problem DescriptionThe 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 is 300.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 x.(it is promised that not all personin 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 as 0.3∗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.InputThere 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 integers n 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,onlycontain 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.OutputFor each testcase, output "The member list of Shandong team is as follows:" without Quotation marks.Followed m lines,every line is the name of the team with their Ultimate mark decreasing.Sample Input210 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 OutputThe 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 14.40解题思路:这道题就是一个简单模拟问题，只要输出前m-1个人就行，最后一个加特判，判断是否有女生并且输出与否，如果还没输出就输出一个女生，否则就输出一个男生。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 110;typedef struct it{char name[100];int sex;double r1,r2;double grade;}It;It itv[maxn];bool cmp(const It& lhs,const It& rhs){return lhs.grade >= rhs.grade;}int main(){int t,n,m;scanf("%d",&t);while(t--){scanf("%d %d",&n,&m);int flag = 1;for(int i=0;i<n;i++){char na[maxn];char se[maxn];double r1,r2;scanf("%s %s %lf %lf",na,se,&r1,&r2);strcpy(itv[i].name,na);if(se[0] == 'm') {itv[i].sex = 1;}else  {itv[i].sex = 0;flag = 0;}//itv[i].grade = r1*0.3 + r2*0.7;itv[i].r1 = r1;itv[i].r2 = r2;}double maxn1 = 0,maxn2 = 0;for(int i=0;i<n;i++){if(maxn1<itv[i].r1)maxn1 = itv[i].r1;if(maxn2<itv[i].r2)maxn2 = itv[i].r2;}for(int i=0;i<n;i++){itv[i].r1 *= (300.0/maxn1);itv[i].r2 *= (300.0/maxn2);itv[i].grade = itv[i].r1*0.3 + itv[i].r2*0.7;}sort(itv,itv+n,cmp);int sign = 1;printf("The member list of Shandong team is as follows:\n");for(int i=0;i<m-1;i++){if(itv[i].sex == 0)sign = 0;puts(itv[i].name);}if(flag == 0 && sign == 1){for(int i=m-1;i<n;i++){if(itv[i].sex == 0){puts(itv[i].name);break;}}}elseputs(itv[m-1].name);}}

Reorder the BooksTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 122 Accepted Submission(s): 91Problem Descriptiondxy has a collection of a series of books called "The Stories of SDOI",There are n(n≤19) books in this series.Every book has a number from 1 to n.dxy puts these books in a book stack with the order of their numbers increasing from top to bottom. dxy takes great care of these books and no one is allowed to touch them.One day Evensgn visited dxy's home, because dxy was dating with his girlfriend, dxy let Evensgn stay at home himself. Evensgn was curious about this series of books.So he took a look at them. He found out there was a story about "Little E&Little Q". While losinghimself in the story,he disrupted the order of the books.Knowing that dxy would be back soon,Evensgn needed to get the books ordered again.But because the books were too heavy.The only thing Evensgn could do was to take out a book from the book stack and and put it at the stack top.Give you the order of the disordered books.Could you calculate the minimum steps Evensgn would use to reorder the books? If you could solve the problem for him,he will give you a signed book "The Stories of SDOI 9: The Story of Little E" as a gift.InputThere are several testcases.There is an positive integer T(T≤30) in the first line standing for the number of testcases.For each testcase, there is an positive integer n in the first line standing for the number of books in this series.Followed n positive integers separated by space standing for the order of the disordered books,the ith integer stands for the ith book's number(from top to bottom).Hint:For the first testcase:Moving in the order of book3,book2,book1 ,(4,1,2,3)→(3,4,1,2)→(2,3,4,1)→(1,2,3,4),and this is the best way to reorder the books.For the second testcase:It's already ordered so there is no operation needed.OutputFor each testcase,output one line for an integer standing for the minimum steps Evensgn would use to reorder the books.Sample Input244 1 2 351 2 3 4 5Sample Output30解题思路:本题比较绕，首先我们可以观察发现如果移动一个数k，那么小于这个数k的数都要移动，那么找到这个k就行。

#include <iostream>#include <cstdio>using namespace std;int num[20];int main(){int t,n;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&num[i]);int m = n;for(int i=n;i>0;i--){if(num[i] == m)m--;}printf("%d\n",m);}return 0;}