1001. A+B Format (20)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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
解:简单的运算,当时考虑的是从左往右每三个一组,有几个测试样例一直不过,正确的做法是从右往左每三个一组,举个例子,如2222,按标准格式输出应该是2,222,而不是222,2。其他的倒是没什么该注意的,数组保存每位的值,输出控制下即可。
贴上代码:
#include<iostream> #include<cmath> #include<cstdlib> #include<cstring> #include<cstdio> using namespace std; int main() { int a,b; while(scanf("%d%d",&a,&b)==2) { int c=a+b; int flag=0; if(c<0) flag=1; if(abs(c)/1000) { int a[10],cnt=0; string s; memset(a,0,sizeof(a)); c=abs(c); do { a[cnt++]=c%10; c=c/10; }while(c); if(flag==1) cout<<"-"; int count=0; for(int i=cnt-1;i>=0;i--) { if((i+1)%3==0&&i!=cnt-1) cout<<','; count++; cout<<a[i]; } cout<<endl; } else { printf("%d ",c); } } return 0; }
版权声明:本文为博主原创文章,未经博主允许不得转载。