2012湖南大学第八届程序设计竞赛 Incredible[公式]
题目地址:http://acm.hnu.cn/online/?action=problem&type=show&id=12312&courseid=215
Incredible |
Time Limit: 2000ms, Special Time Limit:5000ms, Memory Limit:65536KB |
Total submit users: 9, Accepted users: 8 |
Problem 12312 : No special judgement |
Problem description |
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Input |
There are several test cases end with EOF. For each test case, the first line is three integers n ( 0 < n ≤ 50 ), k ( 0 < k ≤ n ) and A ( 0 < A < 10000 ), which are described as above. |
Output |
For each the case, just output a real number rounded to two decimal places which means the least score that one team can be promoted. |
Sample Input |
4 2 3 3 1 2 |
Sample Output |
7.50 4.00 |
Problem Source |
The 2012 8th Hunan University Programming Contest |
Submit Discuss Judge Status Problems Ranklist |
解体报告:
题目大意:
给出N个球队,球队之间两两会进行一场比赛,赢的球队得A分,输的球队不得分,两个球队打平的话各得A/2分。现在在最开始一场球赛都没有打过的前提下,问球队需要至少多少分使得在任意情况下该球队都能排在前K名之内。
code:
1 #include<iostream> 2 using namespace std; 3 int main() 4 { 5 int n,k; 6 double A; 7 double score; 8 while(~scanf("%d%d%lf",&n,&k,&A)) 9 { 10 if(k==n) 11 score=0; 12 else 13 { 14 score=(n-1)*A; 15 while(--k) 16 score-=(A/2.0); 17 } 18 printf("%.2f\n",score); 19 } 20 return 0; 21 }
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Brad(Bowen) Xu
E-Mail : maxxbw1992@gmail.com