1001. A+B Format (20)
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
Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.
Output
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 9Sample Output
-999,991
以下代码有一个案例未通过
#include<stdio.h>
int main()
{
int a,b;
while(scanf("%d%d",&a,&b)!=EOF)
{
int sign=0;
int c=a+b;
if(c<0)
{
c=0-c;
sign=1;
}
int ans=0;
char s[100];
while(c!=0)
{
if(ans%4==0)
s[ans++]=',';
else
{
s[ans++]=c%10+'0';
c/=10;
}
}
if(sign==1)
printf("-");
for(int i=ans-1;i>=1;i--)
printf("%c",s[i]);
if(s[0]!=',')
printf("%c",s[0]);
printf("\n");
}
return 0;
}
经过我查看发现上述代码中只做到了sum非零的情况,而当sum=0时不成立,修改如下,以下这个就通过了:
#include<stdio.h>
void show(int sum){
char str[100];
int k=0;
int ans=0;
if(sum==0){
printf("0\n");
return ;
}
while(sum!=0){
if(k==3){
str[ans++] = ',';
k = 0;
}
else{
k++;
str[ans++]=(sum%10)+'0';
sum = sum/10;
}
}
for(int i=ans-1;i>=0;i--){
printf("%c",str[i]);
}
printf("\n");
}
int main(){
int a,b,sum=0;
while(scanf("%d %d",&a,&b)!=EOF){
sum=a+b;
if(sum>=0){
show(sum);
}
else {
sum=-sum;
printf("-");
show(sum);
}
}
}
以下为他人代码,第一个同理,后一个界定数值范围内可行:
#include<iostream>
#include<cmath>
using namespace std;
int main(){
long a,b,c;
int i,flag;
bool is_positive;
char result[9];
while(cin>>a>>b)
{
c = a+b;
if( c>0 )
is_positive = true;
else
is_positive = false;
flag = 0;
if(abs(c) >= 1000){
c = abs(c); //取绝对值
for( i=0; c>0; i++){
result[i] = c%10 + '0'; //数字转换成字符
c = c/10;
if(flag == 2 && c != 0 )
{
i++;
result[i] = ','; //每三个加一个逗号
flag = 0;
}
else
flag++;
}
if( !is_positive ) //结果为负数时
{
result[i] = '-';
cout<<result[i];
}
i--; //i指向第一个数字
for(; i>=0; i--){
cout<<result[i];
}
cout<<endl;
}
else
{
cout<<c<<endl;
}
}
return 0;
}
网上一般方法还有这个:
#include<stdio.h>
int main()
{
int a,b;
int sum;
while(scanf("%d%d\n",&a,&b) != EOF){
sum = a+b;
if(sum < 0){
printf("-");
sum = -sum;
}
if(sum>=1000000){
printf("%d,%03d,%03d\n",sum/1000000, (sum/1000)%1000, sum%1000);
}
else if(sum >= 1000){
printf("%d,%03d\n",sum/1000,sum%1000);
} else{
printf("%d\n", sum);
}
}
return 0;
}