zoukankan      html  css  js  c++  java
  • hdu 1212 Big Number

    Big Number

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 3386    Accepted Submission(s): 2305


    Problem Description
    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.
     
    Input
    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.
     
    Output
    For each test case, you have to ouput the result of A mod B.
     
    Sample Input
    2 3 12 7 152455856554521 3250
     
    Sample Output
    2 5 1521
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstdlib>
     4 #include <cmath>
     5 #include <cstring>
     6 using namespace std;
     7 char a[1000+10]; int b;
     8 int main(void){
     9 #ifndef ONLINE_JUDGE
    10   freopen("1212.in", "r", stdin);
    11 #endif
    12   while (~scanf("%s%d", a, &b)){
    13     int len = strlen(a); int x = a[len-1]-'0', po = 1;
    14     x = x % b; int y;
    15     for (int i = len - 2; i >= 0; --i){
    16       y = a[i] - '0'; po = (po * 10) % b;
    17       x = (x + y*po) % b;
    18     }
    19     printf("%d\n", x);
    20   }
    21   return 0;
    22 }

    看了讨论版学会的,不过还是有一点儿小问题,学一下数论吧……

  • 相关阅读:
    register based 和 stack based虚拟机的区别
    Java in a Nutshell学习笔记
    Java中interface和abstract class的区别和联系
    Java中final的作用
    Android 源码下载
    Android Fragment 你应该知道的一切
    Android Fragment 真正的完全解析(下)
    Android Fragment 真正的完全解析(上)
    IntelliJ IDEA 使用总结
    Linux在目录中查找某个函数
  • 原文地址:https://www.cnblogs.com/liuxueyang/p/2956561.html
Copyright © 2011-2022 走看看