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/********************************* *    日期：2012-6-23 *    作者：SJF0115 *    题号: 九度1002 *    题目：Grading *    结果：AC *    题意： *    总结：**********************************/#include
   
    #include
    
     int main(){	int P,T,G1,G2,G3,GJ;	double Grade;	while(scanf("%d %d %d %d %d %d",&P,&T,&G1,&G2,&G3,&GJ)!=EOF){		if((G1>P||G1<0)||(G2>P||G2<0)||(G3>P||G3<0)||(GJ>P||GJ<0)){			break;		}		if(fabs(G1-G2)<=T){			Grade = (double)(G1 + G2)/2.0;		}		else{			//G3 is within the tolerance with both G1 and G2			if(fabs(G1-G3)<=T&&fabs(G2-G3)<=T){				if(G1>G2){					if(G1>G3){						Grade = G1;					}					else{						Grade = G3;					}				}				else{					if(G2>G3){						Grade = G2;					}					else{						Grade = G3;					}				}			}			//G3 is within the tolerance with neither G1 nor G2			else if(fabs(G1-G3)>T&&fabs(G2-G3)>T){				Grade = GJ;			}			//G3 is within the tolerance with either G1 or G2, but NOT both			else{				if(fabs(G3 - G1) > fabs(G3 - G2)){					Grade = (double)(G2 + G3)/2.0;				}				else{					Grade = (double)(G1 + G3)/2.0;				}			}		}		printf("%.1lf\n",Grade);	}	return 0;}
    
   

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