PAT 1054. 求平均值
本题的基本要求非常简单:给定N个实数,计算它们的平均值。但复杂的是有些输入数据可能是非法的。一个“合法”的输入是[-1000,1000]区间内的实数,并且最多精确到小数点后2位。当你计算平均值的时候,不能把那些非法的数据算在内。
输入格式:
输入第一行给出正整数N(<=100)。随后一行给出N个实数,数字间以一个空格分隔。
输出格式:
对每个非法输入,在一行中输出“ERROR: X is not a legal number”,其中X是输入。最后在一行中输出结果:“The average of K numbers is Y”,其中K是合法输入的个数,Y是它们的平均值,精确到小数点后2位。如果平均值无法计算,则用“Undefined”替换Y。如果K为1,则输出“The average of 1 number is Y”。
输入样例1:
7
5 -3.2 aaa 9999 2.3.4 7.123 2.35
输出样例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
输入样例2:
2
aaa -9999
输出样例2:
ERROR: aaa is not a legal number
ERROR: -9999 is not a legal number
The average of 0 numbers is Undefined
分析
我用的是抛出异常,下面也附上另一种方式
代码如下
#include<iostream>
#include<stdexcept>
using namespace std;
int main(){
int N,cnt=0;
double sum=0.0;
cin>>N; string s="";
char ch[50];
for(int i=0;i<N;i++){
cin>>s;
try{
sprintf(ch,"%.2lf",stod(s));
for(int j=0;j<s.size();j++)
if(s[j]!=ch[j]) throw invalid_argument("");
if(stod(s)<-1000||stod(s)>1000) throw invalid_argument("");
else {sum+=stod(s);cnt++;}
}catch(invalid_argument err){
cout<<"ERROR: "<<s<<" is not a legal number"<<endl;
}
}
if(cnt == 1) {
printf("The average of 1 number is %.2lf", sum);
} else if(cnt > 1) {
printf("The average of %d numbers is %.2lf", cnt, sum / cnt);
} else {
printf("The average of 0 numbers is Undefined");
}
}
#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;
int main() {
int n, cnt = 0;
char a[50], b[50];
double temp, sum = 0.0;
cin >> n;
for(int i = 0; i < n; i++) {
scanf("%s", a);
sscanf(a, "%lf", &temp);
sprintf(b, "%.2lf",temp);
int flag = 0;
for(int j = 0; j < strlen(a); j++) {
if(a[j] != b[j]) flag = 1;
}
if(flag || temp < -1000 || temp > 1000) {
printf("ERROR: %s is not a legal number\n", a);
continue;
} else {
sum += temp;
cnt++;
}
}
if(cnt == 1) {
printf("The average of 1 number is %.2lf", sum);
} else if(cnt > 1) {
printf("The average of %d numbers is %.2lf", cnt, sum / cnt);
} else {
printf("The average of 0 numbers is Undefined");
}
return 0;
}