Programmer, Rank Thyself
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 1174 | Accepted: 448 |
Description
Implement a ranking program similar to the one used for this programming contest.
Input
The input file contains one or more contests followed by a line containing only zero that signals the end of the file. Each contest begins with a line containing a positive integer c no greater than 20 indicating the number of teams in the contest, and is followed by c lines that contain a team name and the solution times for seven problems, separated by spaces. Team names consist of between one and ten letters. All team names within a contest are unique. Times are nonnegative integers no greater than 500.
As described in the Notes to Teams, teams are ranked first by greatest number of problems solved, then by least total time, then by least geometric mean of all nonzero times. Teams that are still tied are given the same numeric ranking and are listed in alphabetical order, using case-sensitive string comparison. The numeric ranking for a team is always one more than the number of teams ranked ahead of (not tied with) that team. For this problem all geometric means will be rounded to an integer as described below, and only the rounded value will be used when computing rankings and displaying results.
If all times are zero, then the geometric mean is also zero. Otherwise, if there are n nonzero times t1, ..., tn, then the geometric mean is defined to be
exp ((ln t1 + ln t2 + ... + ln tn) / n)
where exp x means ex and ln x means the natural logarithm of x. (There are other mathematically equivalent definitions of the geometric mean, but they may produce slightly different answers due to roundoff and/or overflow problems. Use this definition to get the same answers as the judges.) After computing the geometric mean, round it to an integer by adding 0.5 and truncating any fractional digits. (C/C++ and Java automatically truncate fractions when casting a floating-point type to an integral type. In Pascal use the trunc function.)
As described in the Notes to Teams, teams are ranked first by greatest number of problems solved, then by least total time, then by least geometric mean of all nonzero times. Teams that are still tied are given the same numeric ranking and are listed in alphabetical order, using case-sensitive string comparison. The numeric ranking for a team is always one more than the number of teams ranked ahead of (not tied with) that team. For this problem all geometric means will be rounded to an integer as described below, and only the rounded value will be used when computing rankings and displaying results.
If all times are zero, then the geometric mean is also zero. Otherwise, if there are n nonzero times t1, ..., tn, then the geometric mean is defined to be
exp ((ln t1 + ln t2 + ... + ln tn) / n)
where exp x means ex and ln x means the natural logarithm of x. (There are other mathematically equivalent definitions of the geometric mean, but they may produce slightly different answers due to roundoff and/or overflow problems. Use this definition to get the same answers as the judges.) After computing the geometric mean, round it to an integer by adding 0.5 and truncating any fractional digits. (C/C++ and Java automatically truncate fractions when casting a floating-point type to an integral type. In Pascal use the trunc function.)
Output
For each contest you must output the rankings in a table. All tables will have the same width, with the length equal to the number of teams entered in the contest. Use the exact format shown in the examples. Each contest has a numbered header followed by a table with one team entry per line. Each entry contains the ranking, team name, problems solved, total time, geometric mean, and then the individual solution times in the same order they appeared in the input. In the first example all values completely fill their fields. In general you may need to pad values with spaces (never tabs) so that they have the correct field width. The team name is left-justified, and all other fields are right-justified. The ranking always has two digits, including a leading zero if necessary. Spaces will never appear at the beginning or end of lines.
Sample Input
1 Plutonians 123 234 345 456 167 278 389 4 Xap 0 0 0 0 0 0 0 Foo 20 30 0 50 40 0 10 Bar 0 50 20 0 10 40 30 Baz 0 0 0 0 0 0 0 3 Venus 213 0 0 57 0 0 0 Neptune 0 0 0 117 153 0 0 Mars 0 150 0 0 0 0 120 0
Sample Output
CONTEST 1 01 Plutonians 7 1992 261 123 234 345 456 167 278 389 CONTEST 2 01 Bar 5 150 26 0 50 20 0 10 40 30 01 Foo 5 150 26 20 30 0 50 40 0 10 03 Baz 0 0 0 0 0 0 0 0 0 0 03 Xap 0 0 0 0 0 0 0 0 0 0 CONTEST 3 01 Venus 2 270 110 213 0 0 57 0 0 0 02 Mars 2 270 134 0 150 0 0 0 0 120 02 Neptune 2 270 134 0 0 0 117 153 0 0
Source
c++代码：
2 #include<stdio.h> 3 #include<string.h> 4 5 #include<math.h> 6 7 #include<stdlib.h> 8 9 struct rank 10 11 { 12 13 char name[12]; 14 15 int sol, tot, g, p[7], ind;//sol 解决题数 tot总时间 16 17 }; 18 19 rank r[25]; 20 21 int comp(const void *c, const void *d) 22 23 { 24 25 rank *a = (rank *)c; rank *b = (rank *)d; 26 if(a.sol>b.sol)return 1; 27 else if(a.sol<b.sol)return -1; 28 else 29 { 30 31 if(a.tot<b.tot)return -1; 32 33 else if(a.tot>b.tot) return 1; 34 35 else 36 37 { 38 39 if(a.g>b.g) return 1; 40 41 else if(a.g<b.g)return -1; 42 43 else 44 45 { 46 47 return strcmp(a.name,b.name); 48 49 } 50 51 } 52 53 } 54 55 } 56 57 int main() 58 59 { 60 int n,cnt=1; 61 62 scanf("%d"，&n); 63 64 if(!n) break; 65 66 for(int i=0;i<n;i++) 67 68 { 69 70 r[i].sol=0;r[i].tot=0;r[i].g=0; 71 72 double t=0; 73 74 scanf("%s",r[i].name); 75 76 for(int i=0;i<7;i++) 77 78 { 79 80 scanf("%d",&r[i].p[j]); 81 82 if(r[i].p[j]>0) 83 84 { 85 86 r[i].sol++;r[i].tot+=r[i].p[j]; 87 88 t += log((double)ranks[i].p[j]); 89 90 } 91 92 } 93 94 if(r[i].sol) 95 96 { 97 98 r[i].g=(int)(exp(t/ranks[i].sol) + 0.5); 99 100 } 101 102 } 103 104 qsort(r,n,sizeof(rank),comp); 105 106 printf("CONTEST %d ", cnt); 107 108 cnt++; 109 110 for(int s= 0; s < n; s++) 111 112 { 113 114 if(s < 9) 115 116 { 117 118 printf("0"); 119 120 } 121 122 if(s > 0 && ranks[s].sol == ranks[s-1].sol && ranks[s].tot == ranks[s-1].tot && ranks[s].g == ranks[s-1].g) 123 124 { 125 126 ranks[s].ind = ranks[s-1].ind; 127 128 } 129 130 else 131 132 { 133 134 ranks[s].ind = s + 1; 135 136 } 137 138 printf("%d %-10s %d %4d %3d %3d %3d %3d %3d %3d %3d %3d ", ranks[s].ind, ranks[s].name, ranks[s].sol, ranks[s].tot, ranks[s].g, 139 140 ranks[s].p[0], ranks[s].p[1], ranks[s].p[2], ranks[s].p[3], ranks[s].p[4], ranks[s].p[5], ranks[s].p[6]); 141 142 } 143 144 } 145 146 return 0; 147 148 } 149 150
总结：
1.本题参赛队员的数据过多，建议采用结构体数据类型；
2.采用qsort函数进行快速排序，其中的comp函数应当仔细研究；
3.输出格式应该十分小心，一失足成千古恨啊~。