1108. Finding Average (20)段错误

The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A “legal” input is a real number in [-1000, 1000] and is accurate up to no more than 2 decimal places. When you calculate the average, those illegal numbers must not be counted in.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=100). Then N numbers are given in the next line, separated by one space.

Output Specification:

For each illegal input number, print in a line “ERROR: X is not a legal number” where X is the input. Then finally print in a line the result: “The average of K numbers is Y” where K is the number of legal inputs and Y is their average, accurate to 2 decimal places. In case the average cannot be calculated, output “Undefined” instead of Y. In case K is only 1, output “The average of 1 number is Y” instead.

Sample Input 1:

7

5 -3.2 aaa 9999 2.3.4 7.123 2.35

Sample Output 1:

ERROR: aaa is not a legal number

ERROR: 9999 is not a legal number

ERROR: 2.3.4 is not a legal number

ERROR: 7.123 is not a legal number

The average of 3 numbers is 1.38

Sample Input 2:

2

aaa -9999

Sample Output 2:

ERROR: aaa is not a legal number

ERROR: -9999 is not a legal number

The average of 0 numbers is Undefined

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
void printError(char temp[])
{
printf("ERROR: %s is not a legal number\n",temp);
}
void printOne(int x,float y)
{

printf("The average of %d number is %0.2f\n",x,y);

}
void printMore(int x,float y)
{

printf("The average of %d numbers is %0.2f\n",x,y);

}
void printZero()
{

printf("The average of 0 numbers is Undefined\n");

}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
//是不是合法的
// 小数点后两位
char temp[100];// 原来是n,改成100后结果正确。
float sum=0;
float num=0;
int ct=0;
int dot=0;
bool legal=true;
int length=0;
int dotlocation=0;
bool p=true;
while(n--)
{
num=0;
memset(temp,0,sizeof(temp));
scanf("%s",temp);
dot=0;
legal=true;
length=strlen(temp);
dotlocation=strlen(temp);
p=true;//正负
for(int i=0;i<strlen(temp);i++)
{

if(temp[i]>='0'&&temp[i]<='9'||temp[i]=='.')
{

if(temp[i]=='.')
{
dot++;
dotlocation=i;
if(dot>1)
{
legal=false;
break;
}

if(i+3<strlen(temp))
{
legal=false;
break;
}
}

}
else if(temp[i]=='-')
{
if(i!=0)
legal=false;

}
else
{
legal=false;
break;
}
}
if(legal)
{
for(int i=0;i<strlen(temp);i++)
{
if(temp[i]=='-')
{
p=false;
continue;
}
if(temp[i]!='.')
{
int j=(temp[i]-'0');
float jj=j*pow(10,dotlocation-i-1);
num+=jj;
}
else
dotlocation++;
}
if(!p)
{
num=-num;
}
if(num>1000||num<-1000)
printError(temp);
else {
sum+=num;
ct++;
}
}
else printError(temp);
//超时
}
//段错误
float i=sum/ct;
if(ct==1)
printOne(ct,i);
else if(ct==0)
printZero();
else printMore(ct,i);

}
return 0;
}

后记

这次为了找到段错误的位置，提交 了很多次，最后找到在while(n–)这个块中，通过减少n的大小，来测试，发现又有案例出现段错误，所以重新查看了所有与n相关的地方，终于找到错误，原来是数组大小设置成n了，这是第二次这样的错误了，谨记谨记谨记。另外这道题做的时间很长，缺少系统的规划，应该养成习惯，提前设计好算法。