问题 E: Jack的A+B

问题 E: Jack的A+B

时间限制: 1 Sec  内存限制: 128 MB
提交: 1996  解决: 601
[提交] [状态] [命题人:jsu_admin]

题目描述

现在有整数a,b，请按西方数字数量级方式输出a+b

输入

题目有多组测试数据
每组输入两个整数a，b
(0<=a,b<=10000000)

输出

输出西方数字数量级的a+b

样例输入 Copy

999 1
36 30
100000 100

样例输出 Copy

1,000
66
100,100

提示

输出的数从最低位起，每三位用逗号隔开

 

倒着模拟题意即可，每三位输出一个逗号，注意和为   0 的情况

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
 void int2str(int n, char *str)    {
        char buf[10] = "";
        int i = 0;
        int len = 0;
        int temp = n < 0 ? -n: n;  // temp为n的绝对值
 
       if (str == NULL)
       {
           return;
      }
      while(temp)
       {
           buf[i++] = (temp % 10) + '0';  //把temp的每一位上的数存入buf
           temp = temp / 10;
       }
  
       len = n < 0 ? ++i: i;  //如果n是负数，则多需要一位来存储负号
       str[i] = 0;            //末尾是结束符0
       while(1)
       {
           i--;
           if (buf[len-i-1] ==0)
           {
               break;
           }
           str[i] = buf[len-i-1];  //把buf数组里的字符拷到字符串
       }
       if (i == 0 )
       {
           str[i] = '-';          //如果是负数，添加一个负号
      }
   }

int main(){
    int a,b;
    int sum;
    char num[1000];
    while(scanf("%d%d",&a,&b)!=EOF){
    memset(num,0,sizeof(num));
    sum = a+b;
    if(sum ==0)
    {
        printf("0
");
        continue;
    }
    int k =0;
    int2str(sum,num);
    //printf("%s",num); 
    int h = strlen(num)%3;
    if(h!=0){
    
    for(int i= 0;i<h;i++)
    {
        printf("%c",num[i]);
    }
    if(strlen(num)>3)
    printf(",");
    }
//    int k = 0;
    for(int i = h;i<strlen(num);i++)
    {
        
        printf("%c",num[i]);
        k++;
        if(k%3==0&&i!=strlen(num)-1&&k!=strlen(num)){
            printf(",");
        }
    }
    printf("
");
}
}

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=10000;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,q;
 
int main()
{
#ifdef DEBUG
    freopen("input.in", "r", stdin);
    //freopen("output.out", "w", stdout);
#endif
    while(scanf("%d%d",&n,&m)!=EOF){
        t=n+m;
        string ans;
        k=0;
        while(t){
            k++;
            q=t%10;
            ans=(char)('0'+q)+ans;
            t/=10;
            if(t&&k>=3){
                k=0;
                ans=','+ans;
            }
        }
        if(n+m<0)
            ans='-'+ans;
        if(n+m==0)
            ans='0'+ans;
        cout<<ans<<endl;
    }
 
    //cout << "Hello world!" << endl;
    return 0;
}