版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/github_38201918/article/details/86522534
因为最近在刷题库,想想就把本人可以想到的解法写到博客里,作为整理归纳。未必是最优解,还请各位高手多多包涵,能够指点指点。
题目要求
1001 A+B Format (20 分)
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 Specification:
Each input file contains one test case. Each case contains a pair of integers a and b where −10
6
≤a,b≤10
6
. The numbers are separated by a space.
Output Specification:
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 9
Sample Output:
-999,991
解题思路
回头 补充吧。。
注意事项
回头再说。。。
代码部分
#include<stdlib.h>
#include<stdio.h>
main(){
long int a ,b,c;
scanf("%ld %ld",&a,&b );
c=a+b;
int i;
if(abs(c)>=1000&&abs(c)<1000000){
printf("%ld,",c/1000);
i=abs(c%1000);
for(int t=3;t>0;t--){
printf("%d",i/100);
i=i%100;
i=i*10;
}
}
else if(abs(c)>=1000000){
printf("%ld,",c/1000000);
i=abs(c%1000000/1000);
for(int t=3;t>0;t--){
printf("%d",i/100);
i=i%100;
i=i*10;
}
printf(",");
i=abs(c%1000);
for(int t=3;t>0;t--){
printf("%d",i/100);
i=i%100;
i=i*10;
}
}
else
printf("%ld",c);
}
运行结果