1001 A+B Format
1001 A+B Format
Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).
Input Specification:
Each input file contains one test case. Each case contains a pair of integers a and b where −106≤a,b≤106. The numbers are separated by a space.
Output Specification:
For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.
Sample Input:
-1000000 9
Sample Output:
-999,991
思路:
当输出多行数值时,一定要注意是否有多余或缺少空行的情况
将 int 转换为 string 可以直接使用 to_string(x)
使用string 代替char数组的好处是 可以使用string自带的函数简化代码,提高效率。
代码:
#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int main(){
int a,b,sum,t=0,l;
char s[100];
while(scanf("%d%d",&a,&b) != -1){
sum = a + b;
sprintf(s, "%d", sum);
if(s[0] == '-'){
if(strlen(s) < 5){
cout<<s;
}else{
cout<<s[0];
l = strlen(s) - 1;
for(int i=1; i<=l % 3; i++){
cout<<s[i];
}
if(l % 3 != 0) cout<<",";
t = 0;
for(int i=l % 3 + 1; i < strlen(s); i++){
cout<<s[i];
t++;
if(t == 3 && i != strlen(s) - 1){
cout<<",";
t = 0;
}
}
}
}
else{
if(strlen(s) < 4){
cout<<s;
}else{
l = strlen(s);
for(int i=0; i<l % 3; i++){
cout<<s[i];
}
if(l % 3 != 0) cout<<",";
t = 0;
for(int i=l % 3; i < strlen(s); i++){
cout<<s[i];
t++;
if(t == 3 && i != strlen(s) - 1){
cout<<",";
t = 0;
}
}
}
}
cout<<endl;
}
return 0;
}
简化代码:
#include <iostream>
using namespace std;
int main() {
int a, b;
cin >> a >> b; //输入a和b
string s = to_string(a + b); //将a+b的值转换为字符串
if(s[0] == '-') { //处理符号
cout << '-';
s.erase(0, 1);
}
int count = 0; //用于记录当前位置
for(int i = s.length() - 1; i >= 0; i--){ //添加逗号
count++;
if(count % 3 == 0 && i > 0){
s.insert(i, ",");
}
}
cout << s;
}

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