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<<","; } }