hdoj1212
2015-08-06 21:18
330 查看
Big Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6010 Accepted Submission(s): 4201
[align=left]Problem Description[/align]
As we know, Big Number is always troublesome. But it's really important in our ACM. And today, your task is to write a program to calculate A mod B.
To make the problem easier, I promise that B will be smaller than 100000.
Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.
[align=left]Input[/align]
The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000, and B will be smaller than 100000. Process to the end of file.
[align=left]Output[/align]
For each test case, you have to ouput the result of A mod B.
[align=left]Sample Input[/align]
2 3 12 7 152455856554521 3250
[align=left]Sample Output[/align]
2 5 1521 [code]# include <stdio.h> int n ; char str[1010] ; int calc() { int i, cc = 0 ; for (i = 0 ; str[i] ; i++) { cc = cc * 10 + str[i] - '0' ; cc %= n ; } return cc ; } int main () { while (~scanf ("%s %d%*c", &str, &n)) printf ("%d\n", calc()) ; return 0 ; }
[/code]
相关文章推荐
- 【西祠日志】【17】初识AngularJS,下一代Web应用的前端
- 妙味课堂视频笔记总结
- 2.Kali安装VMware tools(详细+异常处理)
- iOS设计模式 - 单例
- python练习之通过python pexpect实现自动生成openssl证书
- poj2386水洼dfs
- UDK控制台命令概览
- 十一讲,买花
- LeetCode(66)题解: Plus One
- (2015多校第6场)HDU5361--In Touch (Dijkstra应用)
- [leedcode 216] Combination Sum III
- 测试用例设计白皮书--测试用例设计综合策略
- uva 12307(点集的外接矩形)
- java设计模式(9) - 适配器模式
- Zoj 1671 Walking Ant(BFS+优先队列||记忆化搜索)
- 测试用例设计白皮书--场景设计方法
- 排序小试牛刀(类冒泡……)
- ajax访问不到服务端静态资源文件的处理方法
- Zoj 2100 Seeding
- java中进行二进制,八进制,十六进制,十进制间进行相互转换