九度OJ 1002:Grading
2015-10-14 22:08
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时间限制:1 秒
内存限制:32 兆
特殊判题:否
提交:18410
解决:4753
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:
来源:2011年浙江大学计算机及软件工程研究生机试真题
思路:
基本的单个数据处理题,主要考察分支判断代码撰写。
代码:
内存限制:32 兆
特殊判题:否
提交:18410
解决:4753
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:
20 2 15 13 10 18
样例输出:
14.0
来源:2011年浙江大学计算机及软件工程研究生机试真题
思路:
基本的单个数据处理题,主要考察分支判断代码撰写。
代码:
#include <stdio.h> #include <stdlib.h> #define N 1000 double t; double aver(double a, double b) { return (a+b)/2; } int tol(double i, double j) { return abs(i-j)<=t; } double max(double a, double b, double c) { a = (a>b) ? a : b; a = (a>c) ? a : c; return a; } int main(void) { double p, g1, g2, g3, gj; double score; while (scanf("%lf", &p) != EOF) { scanf("%lf%lf%lf%lf%lf", &t, &g1, &g2, &g3, &gj); if (tol(g1, g2)) score = aver(g1, g2); else if (tol(g1, g3) && tol(g2, g3)) score = max(g1, g2, g3); else if (tol(g1, g3)) score = aver(g1, g3); else if (tol(g2, g3)) score = aver(g2, g3); else score = gj; printf("%.1lf\n", score); } return 0; } /************************************************************** Problem: 1002 User: liangrx06 Language: C Result: Accepted Time:0 ms Memory:912 kb ****************************************************************/
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