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[] = "";
         int i = ;
         int len = ;
         int temp = n <  ? -n: n;  // temp为n的绝对值

        if (str == NULL)
        {
            return;
       }
       while(temp)
        {
            buf[i++] = (temp % ) + '';  //把temp的每一位上的数存入buf
            temp = temp / ;
        }

        len = n <  ? ++i: i;  //如果n是负数，则多需要一位来存储负号
        str[i] = ;            //末尾是结束符0
        while()
        {
            i--;
            if (buf[len-i-] ==)
            {
                break;
            }
            str[i] = buf[len-i-];  //把buf数组里的字符拷到字符串
        }
        if (i ==  )
        {
            str[i] = '-';          //如果是负数，添加一个负号
       }
    }

 int main(){
     int a,b;
     int sum;
     char num[];
     while(scanf("%d%d",&a,&b)!=EOF){
     memset(num,,sizeof(num));
     sum = a+b;
     if(sum ==)
     {
         printf("0\n");
         continue;
     }
     int k =;
     int2str(sum,num);
     //printf("%s",num);
     int h = strlen(num)%;
     if(h!=){

     for(int i= ;i<h;i++)
     {
         printf("%c",num[i]);
     }
     if(strlen(num)>)
     printf(",");
     }
 //    int k = 0;
     for(int i = h;i<strlen(num);i++)
     {

         printf("%c",num[i]);
         k++;
         if(k%==&&i!=strlen(num)-&&k!=strlen(num)){
             printf(",");
         }
     }
     printf("\n");
 }
 }

 /*
 *@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=;
 const int MOD=1e9+;
 const double PI = acos(-1.0);
 const double EXP = 1E-;
 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=;
         while(t){
             k++;
             q=t%;
             ans=(char)(''+q)+ans;
             t/=;
             if(t&&k>=){
                 k=;
                 ans=','+ans;
             }
         }
         if(n+m<)
             ans='-'+ans;
         if(n+m==)
             ans=''+ans;
         cout<<ans<<endl;
     }

     //cout << "Hello world!" << endl;
     return ;
 }