努力总会有收获的，虽然一直在刷水题，但感觉代码的正确率越来越高了。坚持是最重要的，看到几个大神的博客，hdu，poj题目数量上百道，真心佩服，希望我也能坚持下来。
What Is Your Grade?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4032 Accepted Submission(s): 1198
Problem Description
“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<p. A test case starting with a negative integer terminates the input and this test case should not to be processed.
Output
Output the scores of N students in N lines for each case, and there is a blank line after each case.
Sample Input
4 5 06:30:17 4 07:31:27 4 08:12:12 4 05:23:13 1 5 06:30:17 -1
Sample Output
100 90 90 95 100
Author
lcy
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 using namespace std; 5 struct Sa 6 { 7 int m; 8 char t[20]; 9 int sum; 10 int p; 11 }; 12 bool comp(struct Sa a,struct Sa b) 13 { 14 if(a.m!=b.m) 15 return a.m>b.m; //答题数目降序 16 else if(strcmp(a.t,b.t)<0) 17 return true; 18 else return false; 19 } 20 bool comp2(struct Sa a,struct Sa b) 21 { 22 if(a.p<b.p) 23 return true; 24 else return false; 25 } 26 int main() 27 { 28 int n; 29 while(~scanf("%d",&n)&&n!=-1) 30 { 31 struct Sa a[1000]; 32 int i,num4=0,num3=0,num2=0,num1=0,num5=0; 33 for(i=0;i<n;i++) 34 { 35 scanf("%d",&a[i].m); 36 if(a[i].m==5) 37 a[i].sum=100; 38 if(a[i].m==0) 39 a[i].sum=50; 40 scanf("%s",&a[i].t); 41 a[i].p=i; 42 } 43 sort(a,a+n,comp); 44 for(i=0;i<n;i++) 45 { 46 if(a[i].m==5) 47 num5++; 48 if(a[i].m==4) 49 num4++; 50 if(a[i].m==3) 51 num3++; 52 if(a[i].m==2) 53 num2++; 54 if(a[i].m==1) 55 num1++; 56 } 57 for(i=num5;i<num5+num4/2;i++) 58 { 59 a[i].sum=95; 60 } 61 for(i=num5+num4/2;i<num5+num4;i++) 62 { 63 a[i].sum=90; 64 } 65 for(i=num5+num4;i<num5+num4+num3/2;i++) 66 { 67 a[i].sum=85; 68 } 69 for(i=num5+num4+num3/2;i<num5+num4+num3;i++) 70 { 71 a[i].sum=80; 72 } 73 for(i=num5+num4+num3;i<num5+num4+num3+num2/2;i++) 74 { 75 a[i].sum=75; 76 } 77 for(i=num5+num4+num3+num2/2;i<num5+num4+num3+num2;i++) 78 { 79 a[i].sum=70; 80 } 81 for(i=num5+num4+num3+num2;i<num5+num4+num3+num2+num1/2;i++) 82 { 83 a[i].sum=65; 84 } 85 for(i=num5+num4+num3+num2+num1/2;i<num5+num4+num3+num2+num1;i++) 86 { 87 a[i].sum=60; 88 } 89 sort(a,a+n,comp2); 90 for(i=0;i<n;i++) 91 printf("%d ",a[i].sum); 92 printf(" "); 93 } 94 }
代码看起来蛮恐怖的...囧...其实那堆num+什么的可以用个数组表示，在读入数据的时候就统计，用了两次sort(),对于这种需要排序后计算然后按原来顺序输出的情况，就在结构体中建一个标记专门用来存放原来的顺序，第一次排序被打乱后，再排一次序就回来了。strcmp()比较字符串大小非常好用。