2011年浙江大学计算机及软件工程研究生机试真题(2)
2012-04-21 21:18
363 查看
题目描述:
Grading hundreds of thousands of Graduate Entrance Exams is a hard work. It is even harder to design a process to make the results as fair as possible. One way is to assign each exam problem to 3 independent experts. If they do not agree to each other,
a judge is invited to make the final decision. Now you are asked to write a program to help this process.
For each problem, there is a full-mark P and a tolerance T(<P) given. The grading rules are:
• A problem will first be assigned to 2 experts, to obtain G1 and G2. If the difference is within the tolerance, that is, if |G1 - G2| ≤ T, this problem's grade will be the average of G1 and G2.
• If the difference exceeds T, the 3rd expert will give G3.
• If G3 is within the tolerance with either G1 or G2, but NOT both, then this problem's grade will be the average of G3 and the closest grade.
• If G3 is within the tolerance with both G1 and G2, then this problem's grade will be the maximum of the three grades.
• If G3 is within the tolerance with neither G1 nor G2, a judge will give the final grade GJ.
输入:
Each input file may contain more than one test case.
Each case occupies a line containing six positive integers: P, T, G1, G2, G3, and GJ, as described in the problem. It is guaranteed that all the grades are valid, that is, in the interval [0, P].
输出:
For each test case you should output the final grade of the problem in a line. The answer must be accurate to 1 decimal place.
样例输入:
样例输出:
AC代码:
感觉代码写的很乱,一开始就搞错了一步,虽然给出的例子能运行,但是提交老是WA,但一直没有发现这个错误,一直改来改去,最后认真验证数据时才发现问题所在,懒得去优化了,就这样看吧!
Grading hundreds of thousands of Graduate Entrance Exams is a hard work. It is even harder to design a process to make the results as fair as possible. One way is to assign each exam problem to 3 independent experts. If they do not agree to each other,
a judge is invited to make the final decision. Now you are asked to write a program to help this process.
For each problem, there is a full-mark P and a tolerance T(<P) given. The grading rules are:
• A problem will first be assigned to 2 experts, to obtain G1 and G2. If the difference is within the tolerance, that is, if |G1 - G2| ≤ T, this problem's grade will be the average of G1 and G2.
• If the difference exceeds T, the 3rd expert will give G3.
• If G3 is within the tolerance with either G1 or G2, but NOT both, then this problem's grade will be the average of G3 and the closest grade.
• If G3 is within the tolerance with both G1 and G2, then this problem's grade will be the maximum of the three grades.
• If G3 is within the tolerance with neither G1 nor G2, a judge will give the final grade GJ.
输入:
Each input file may contain more than one test case.
Each case occupies a line containing six positive integers: P, T, G1, G2, G3, and GJ, as described in the problem. It is guaranteed that all the grades are valid, that is, in the interval [0, P].
输出:
For each test case you should output the final grade of the problem in a line. The answer must be accurate to 1 decimal place.
样例输入:
20 2 15 13 10 18
样例输出:
14.0
AC代码:
#include<stdio.h> #include <stdlib.h> //using namespace std; int main(){ int p,t,g1,g2,g3,gj; float max; while(scanf("%d%d%d%d%d%d",&p,&t,&g1,&g2,&g3,&gj)!=EOF){ if(abs(g1-g2)<=t){ max=(float)(g1+g2)/2; printf("%.1f\n",max); continue; } else if(abs(g3-g1)<=t&&abs(g3-g2)<=t) { g3=g3>g1?g3:g1; g3=g3>g2?g3:g2; max=(float)g3; printf("%.1f\n",max); continue; } else if(abs(g3-g1)<=t||abs(g3-g2)<=t){ max=(float)((abs(g3-g1)>abs(g3-g2)?g2:g1)+g3)/2; printf("%.1f\n",max); continue; } else { max=(float)gj; printf("%.1f\n",max); continue; } } return 0; }
感觉代码写的很乱,一开始就搞错了一步,虽然给出的例子能运行,但是提交老是WA,但一直没有发现这个错误,一直改来改去,最后认真验证数据时才发现问题所在,懒得去优化了,就这样看吧!
相关文章推荐
- 2011年浙江大学计算机及软件工程研究生机试真题(3)
- A+B for Matrices(2011年浙江大学计算机及软件工程研究生机试真题)
- 2011年浙江大学计算机及软件工程研究生机试真题
- 1002 Grading(2011年浙江大学计算机及软件工程研究生机试真题)
- 1004 Median(2011年浙江大学计算机及软件工程研究生机试真题)
- 九度 1002- A+B for Matrices -2011年浙江大学计算机及软件工程研究生机试真题
- 2011年浙江大学计算机及软件工程研究生机试真题(1)
- 2011年浙江大学计算机及软件工程研究生机试真题
- 题目1002:Grading 2011年浙江大学计算机及软件工程研究生机试真题
- 题目1004:Median 2011年浙江大学计算机及软件工程研究生机试真题
- 2005年浙江大学计算机及软件工程研究生机试真题 畅通工程
- 2005年浙江大学计算机及软件工程研究生机试真题并查集UnionFindjava实现
- 2005年浙江大学计算机及软件工程研究生机试真题 A + B 九度 1010
- 九度题目1015:还是A+B && 2006年浙江大学计算机及软件工程研究生机试真题
- 九度题目1020:最小长方形 2007年浙江大学计算机及软件工程研究生机试真题
- 题目1027:欧拉回路 2008年浙江大学计算机及软件工程研究生机试真题
- 2008年浙江大学计算机及软件工程研究生机试真题
- 2007年浙江大学计算机及软件工程研究生机试真题
- 题目1007:奥运排序问题 2010年浙江大学计算机及软件工程研究生机试真题
- 题目1012:畅通工程(2005年浙江大学计算机及软件工程研究生机试真题)