- 题目描述：
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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：
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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：
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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：
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20 2 15 13 10 18
- 样例输出：
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14.0
- 代码
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#include <iostream> #include <stdio.h> //#include <cmath> #include <cstdlib> using namespace std; int main() { int P,T,G1,G2,G3,GJ; float GF; while(scanf("%d %d %d %d %d %d",&P,&T,&G1,&G2,&G3,&GJ)!=EOF) { //cout << P << " "<<T<< " "<<G1<< " "<<G2<< " "<<G3<< " "<<GJ << endl; if(abs(G1-G2) <= T) { GF = (float)(G1*1.0 + G2*1.0) /2; printf("%.1lf\n",GF); } else if(abs(G3-G2) <= T && abs(G3-G1) <= T) { float temp; if(G1 > G2) { temp = G1; if(G1 < G3) temp = G3; } else { temp = G2; if(G2 < G3) temp = G3; } printf("%.1lf\n",temp); } else if(abs(G3-G2) <= T || abs(G3-G1) <= T) { if(abs(G3-G2) <= T) { GF = (float)(G3 + G2)/2; printf("%.1lf\n",GF); } if(abs(G3-G1) <= T) { GF = (float)(G3 + G1)/2; printf("%.1lf\n",GF); } } else { GF = (float)GJ; printf("%.1lf\n",GF); } } return 0; }注意：
- printf("%.1lf\n",GF);若GF为整数则输出零 需要强制类型转换