hdu 1.3.3/hed 1084
2016-11-13 20:33
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hdu ACM STEPS 1.3.3
“Point, point, life of student!”
This is a ballad(歌谣)well known in colleges, and you must care about your score in this exam too. How many points can you get? Now, I told you the rules which are used in this course.
There are 5 problems in this final exam. And I will give you 100 points if you can solve all 5 problems; of course, it is fairly difficulty for many of you. If you can solve 4 problems, you can also get a high score 95 or 90 (you can get the former(前者) only when your rank is in the first half of all students who solve 4 problems). Analogically(以此类推), you can get 85、80、75、70、65、60. But you will not pass this exam if you solve nothing problem, and I will mark your score with 50.
Note, only 1 student will get the score 95 when 3 students have solved 4 problems.
I wish you all can pass the exam!
Come on!
Input
Input contains multiple test cases. Each test case contains an integer N (1<=N<=100, the number of students) in a line first, and then N lines follow. Each line contains P (0<=P<=5 number of problems that have been solved) and T(consumed time). You can assume that all data are different when 0
“Point, point, life of student!”
This is a ballad(歌谣)well known in colleges, and you must care about your score in this exam too. How many points can you get? Now, I told you the rules which are used in this course.
There are 5 problems in this final exam. And I will give you 100 points if you can solve all 5 problems; of course, it is fairly difficulty for many of you. If you can solve 4 problems, you can also get a high score 95 or 90 (you can get the former(前者) only when your rank is in the first half of all students who solve 4 problems). Analogically(以此类推), you can get 85、80、75、70、65、60. But you will not pass this exam if you solve nothing problem, and I will mark your score with 50.
Note, only 1 student will get the score 95 when 3 students have solved 4 problems.
I wish you all can pass the exam!
Come on!
Input
Input contains multiple test cases. Each test case contains an integer N (1<=N<=100, the number of students) in a line first, and then N lines follow. Each line contains P (0<=P<=5 number of problems that have been solved) and T(consumed time). You can assume that all data are different when 0
#include<iostream> #include<cstdio> #include<cstring> #include<iomanip> #include<cmath> #include<map> #include<set> #include<queue> #include<cstdio> #include<algorithm> #include<string> #include<sstream> #include<ctime> #define ll long long #define INF 0x3f3f3f3f using namespace std; struct node { int solv,h,m,s,xuhao; } student[104]; bool cmp(node a,node b) { if(a.solv!=b.solv) return a.solv>b.solv; else { if(a.h!=b.h){return a.h<b.h;} else { if(a.m!=b.m){return a.m<b.m;} else {return a.s<b.s;} } } } int grade[103]; int nums[6],cnt[6]; int main() { ios::sync_with_stdio(false); int n; int solve,h,m,s; char ch; while(cin>>n&&(n!=-1)) { memset(nums,0,sizeof nums); memset(cnt,0,sizeof cnt); for(int i=0;i<n;i++){ cin>>student[i].solv>>student[i].h>>ch>>student[i].m>>ch>>student[i].s; nums[student[i].solv]++; student[i].xuhao=i; } for(int i=0;i<6;i++){ nums[i]/=2; } sort(student,student+n,cmp); /*for(int i=0;i<n;i++){ cout<<student[i].solv<<" "<<student[i].h<<ch<<student[i].m<<ch<<student[i].s<<" "<<student[i].xuhao<<endl; }*/ for(int i=0;i<n;i++){ if(student[i].solv==5) solve=5,grade[student[i].xuhao]=100; else if(student[i].solv==0) grade[student[i].xuhao]=50; else{ if(cnt[student[i].solv]<nums[student[i].solv]){ cnt[student[i].solv]++; if(student[i].solv==4) solve=4,grade[student[i].xuhao]=95; if(student[i].solv==3) solve=3,grade[student[i].xuhao]=85; if(student[i].solv==2) solve=2,grade[student[i].xuhao]=75; if(student[i].solv==1) solve=1,grade[student[i].xuhao]=65; } else{ if(student[i].solv==4) grade[student[i].xuhao]=90; if(student[i].solv==3) grade[student[i].xuhao]=80; if(student[i].solv==2) grade[student[i].xuhao]=70; if(student[i].solv==1) grade[student[i].xuhao]=60; } } } for(int i=0;i<n;i++){ cout<<grade[i]<<endl; } cout<<endl; } return 0; }
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