POJ 2680 化学诊断

时间限制: 

1000ms

内存限制: 

65536kB

描述

下表是进行血常规检验的正常值参考范围，及化验值异常的临床意义：


给定一张化验单，判断其所有指标是否正常，如果不正常，统计有几项不正常。化验单上的值必须严格落在正常参考值范围内，才算是正常。正常参考值范围包括边界，即落在边界上也算正常。

输入

输出

对于每组测试数据，输出一行。如果所有检验项目正常，则输出：normal；否则输出不正常的项的数目。

样例输入

2

female 4.5 4.0 115 37 200

male 3.9 3.5 155 36 301

样例输出

normal

3

 

（1）、源代码：

#include <iostream>

#include <cstring>

#include <string>

 

using namespace std;

 

int main()

{

                int i, j;

                int k;

                string gender;

                double wbc, rbc, hgb, hct, plt;

                int num = 0;

 

                cin >> k;

                for(i = 0; i < k; i++)

                {

                                cin >> gender >> wbc >> rbc >> hgb >> hct >> plt;

                                if(wbc < 4.0 || wbc >10.0)

                                                num++;

                                if(rbc < 3.5 || rbc > 5.5)

                                                num++;

                                if(gender == "male")

                                {

                                                if(hgb < 120 || hgb > 160)

                                                                num++;

                                                if(hct < 42 || hct > 48)

                                                                num++;

                                }

                                else if(gender == "female")

                                {

                                                if(hgb < 110 || hgb > 150)

                                                                num++;

                                                if(hct < 36 || hct > 40)

                                                                num++;

                                }

                                if(plt < 100 || plt > 300)

                                                num++;

                                if(num == 0)

                                                cout << "normal" << endl;

                                else

                                                cout << num << endl;

                                num = 0;

                }

                return 0;

}

 

（2）、解题思路：略

（3）、出错原因：略。

 

 

 

 

转载于:https://www.cnblogs.com/lydf-2012/archive/2012/05/02/2479660.html