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2013
12-13

九度-1002-Grading[考研机试]

题目来源:2011年浙江大学计算机及软件工程研究生机试真题

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

模拟题。细心即可。

#include 
#include 
#include 
 
int main(){
    int p,t,g1,g2,g3,gj;
 
 
    while(scanf("%d %d %d %d %d %d",&p,&t,&g1, &g2,&g3, &gj) != EOF){
        double result;
        if(abs(g1 - g2) <= t){
            result = (g1 + g2)/2.0;
        }else{
            int sub1 =t- abs(g3 - g1);
            int sub2 =t- abs(g3 - g2);
            if(sub1 >= 0 && sub2 >= 0){
//              result = (g1+g2+g3) / 3.0;
                int max1 = g1 > g2 ? g1:g2;
                result = max1 > g3 ? max1:g3;
            }
            else if(sub1 >= 0){
                result = (g3 + g1)/2.0;
            }else if(sub2 >= 0){
                result = (g2+g3)/2.0;
            }else if(sub1<0 && sub2 <0){
                result = gj;
            }
        }
 
        printf("%.1lf\n",result);
    }
 
    return 0;
}
/**************************************************************
    Problem: 1002
    User: 从此醉
    Language: C++
    Result: Accepted
    Time:0 ms
    Memory:1020 kb
****************************************************************/

  1. #!/usr/bin/env python
    def cou(n):
    arr =
    i = 1
    while(i<n):
    arr.append(arr[i-1]+selfcount(i))
    i+=1
    return arr[n-1]

    def selfcount(n):
    count = 0
    while(n):
    if n%10 == 1:
    count += 1
    n /= 10
    return count