// 1002_Grading.cpp : 定義控制臺(tái)應(yīng)用程序的入口點(diǎn)。//題目1002：Grading//時(shí)間限制：1 秒內(nèi)存限制：32 兆特殊判題：否提交：21415解決：5522//題目描述：// 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.//輸入：// 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].//輸出：// 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.//樣例輸入：//20 2 15 13 10 18//樣例輸出：//14.0//來(lái)源：//2011年浙江大學(xué)計(jì)算機(jī)及軟件工程研究生機(jī)試真題#include "stdafx.h"#include "stdio.h"#include "math.h"int main(){ int p,t,g1,g2,g3,gj; while(scanf("%d %d %d %d %d %d",&p,&t,&g1,&g2,&g3,&gj)!=EOF) { if(abs(g1-g2)<=t) printf("%.1f/n",(float)(g1+g2)/2); else if(abs(g3-g1)<=t && abs(g3-g2)<=t) { int max = g1; if(g2>g1) max =g2; if(g3>g1&&g3>g2) max= g3; printf("%.1f/n",(float)max); } else if(abs(g3-g1)<=t) printf("%.1f/n",(float)(g3+g1)/2); else if (abs(g3-g2)<=t) printf("%.1f/n",(float)(g3+g2)/2); else if(abs(g3-g2)>t &&abs(g3-g1)>t) printf("%.1f/n",(float)gj); } //printf("%.1f",(float)(3+4)/2); return 0;}/*abs();*/
