A+B Format (20)
时间限制 1000 ms 内存限制 32768 KB 代码长度限制 100 KB 判断程序 Standard (来自 小小)
题目描述
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).
输入描述:
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.
输出描述:
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.
输入例子:
-1000000 9
输出例子:
-999,991
#include <bits/stdc++.h>
#include <stdlib.h>
using namespace std;
#define vi vector<int>
#define pii pair<int,int>
#define x first
#define y second
#define all(x) x.begin(),x.end()
#define pb push_back
#define mp make_pair
#define SZ(x) x.size()
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define pi acos(-1)
#define mod 1000000007
#define inf 1000000007
#define ll long long
#define DBG(x) cerr<<(#x)<<"="<<x<<"\n";
#define N 200010
template <class U,class T> void Max(U &x, T y){if(x<y)x=y;}
template <class U,class T> void Min(U &x, T y){if(x>y)x=y;}
template <class T> void add(int &a,T b){a=(a+b)%mod;}
int main(){
int a,b,ans[10];
scanf("%d %d",&a,&b);
int sum=0,num=0;
sum=a+b;
if(sum<0)
{
sum=-sum;
cout<<'-';
}
while(sum!=0)
{
ans[num++]=sum%10;
sum=sum/10;
}
per(i,0,num)
{
cout<<ans[i];
if(i%3==0&&i>0)cout<<",";
}
}