题目描述
链接
判断是否合法字符串
分析
- 巧解
sscanf(s1, "%lf", &num)
– 将字符串s1中按格式写入变量num
sprintf(s2, "%.2f", num)
– 将变量num按格式写入字符串s2 - 我的笨方法
- 正数和负数判断
- 判断小数点的个数及位置,及小数点后的位数,及数字的大小
stof
的应用
- 一个巨坑的地方是!!!如果只有一个数,那么英文用单数!!!即
number
,而不是numbers
代码1
#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, "%.2f",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
", a);
continue;
} else {
sum += temp;
cnt++;
}
}
if(cnt == 1)
printf("The average of 1 number is %.2f", sum);
else if(cnt > 1)
printf("The average of %d numbers is %.2f", cnt, sum / cnt);
else
printf("The average of 0 numbers is Undefined");
return 0;
}
代码2
#include<bits/stdc++.h>
using namespace std;
bool check(string s){
int len = s.length();
if(s[0] != '-' && (s[0] < '0' || s[0] > '9')) return false; //正负号
int dot_cnt = 0;
int dot_loc = -1;
for(int i=1;i<len;i++){
if(s[i] == '.'){ //统计点数
dot_cnt++;
if(dot_loc == -1) dot_loc = i;
continue;
}
if((s[i] < '0' || s[i] > '9') && s[i] != '.') return false; //是否为非数字字符
}
if(dot_cnt > 1) return false; //小数点个数
else if(dot_cnt == 1){
if(len - 1 - dot_loc > 2) return false; //小数点后位数
}
if(stof(s)<-1000 || stof(s)>1000) return false; //数字范围
return true;
}
int main(){
int n;
string s;
cin>>n;
double sum = 0;
int cnt = 0;
for(int i=0;i<n;i++){
cin>>s;
if(check(s)){
cnt++;
sum += stof(s);
}else{
cout<<"ERROR: "<<s<<" is not a legal number"<<endl;
}
}
if(cnt == 0){
cout<<"The average of 0 numbers is Undefined"<<endl;
}else if(cnt != 1){
printf("The average of %d numbers is %.2f
", cnt, sum/cnt);
}else{
printf("The average of 1 number is %.2f
", sum);
}
}