题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=1212

Big Number

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.

SampleInput

2 3
12 7
152455856554521 3250

SampleOutput

2
5
1521

大数取余模板题。。

 #include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<cstdio>
#include<vector>
#include<string>
#include<map>
#include<set>
using std::cin;
using std::max;
using std::cout;
using std::endl;
using std::string;
using std::istream;
using std::ostream;
#define sz(c) (int)(c).size()
#define all(c) (c).begin(), (c).end()
#define iter(c) decltype((c).begin())
#define cls(arr,val) memset(arr,val,sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define fork(i, k, n) for (int i = (int)k; i <= (int)n; i++)
#define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)
#define pb(e) push_back(e)
#define mp(a, b) make_pair(a, b)
struct BigN {
typedef unsigned long long ull;
static const int Max_N = ;
int len, data[Max_N];
BigN() { memset(data, , sizeof(data)), len = ; }
BigN(const int num) {
memset(data, , sizeof(data));
*this = num;
}
BigN(const char *num) {
memset(data, , sizeof(data));
*this = num;
}
void clear() { len = , memset(data, , sizeof(data)); }
BigN& clean(){ while (len > && !data[len - ]) len--; return *this; }
string str() const {
string res = "";
for (int i = len - ; ~i; i--) res += (char)(data[i] + '');
if (res == "") res = "";
res.reserve();
return res;
}
BigN operator = (const int num) {
int j = , i = num;
do data[j++] = i % ; while (i /= );
len = j;
return *this;
}
BigN operator = (const char *num) {
len = strlen(num);
for (int i = ; i < len; i++) data[i] = num[len - i - ] - '';
return *this;
}
BigN operator + (const BigN &x) const {
BigN res;
int n = max(len, x.len) + ;
for (int i = , g = ; i < n; i++) {
int c = data[i] + x.data[i] + g;
res.data[res.len++] = c % ;
g = c / ;
}
return res.clean();
}
BigN operator * (const BigN &x) const {
BigN res;
int n = x.len;
res.len = n + len;
for (int i = ; i < len; i++) {
for (int j = , g = ; j < n; j++) {
res.data[i + j] += data[i] * x.data[j];
}
}
for (int i = ; i < res.len - ; i++) {
res.data[i + ] += res.data[i] / ;
res.data[i] %= ;
}
return res.clean();
}
BigN operator * (const int num) const {
BigN res;
res.len = len + ;
for (int i = , g = ; i < len; i++) res.data[i] *= num;
for (int i = ; i < res.len - ; i++) {
res.data[i + ] += res.data[i] / ;
res.data[i] %= ;
}
return res.clean();
}
BigN operator - (const BigN &x) const {
assert(x <= *this);
BigN res;
for (int i = , g = ; i < len; i++) {
int c = data[i] - g;
if (i < x.len) c -= x.data[i];
if (c >= ) g = ;
else g = , c += ;
res.data[res.len++] = c;
}
return res.clean();
}
BigN operator / (const BigN &x) const {
BigN res, f = ;
for (int i = len - ; ~i; i--) {
f *= ;
f.data[] = data[i];
while (f >= x) {
f -= x;
res.data[i]++;
}
}
res.len = len;
return res.clean();
}
BigN operator % (const BigN &x) {
BigN res = *this / x;
res = *this - res * x;
return res;
}
BigN operator += (const BigN &x) { return *this = *this + x; }
BigN operator *= (const BigN &x) { return *this = *this * x; }
BigN operator -= (const BigN &x) { return *this = *this - x; }
BigN operator /= (const BigN &x) { return *this = *this / x; }
BigN operator %= (const BigN &x) { return *this = *this % x; }
bool operator < (const BigN &x) const {
if (len != x.len) return len < x.len;
for (int i = len - ; ~i; i--) {
if (data[i] != x.data[i]) return data[i] < x.data[i];
}
return false;
}
bool operator >(const BigN &x) const { return x < *this; }
bool operator<=(const BigN &x) const { return !(x < *this); }
bool operator>=(const BigN &x) const { return !(*this < x); }
bool operator!=(const BigN &x) const { return x < *this || *this < x; }
bool operator==(const BigN &x) const { return !(x < *this) && !(x > *this); }
friend istream& operator >> (istream &in, BigN &x) {
string src;
in >> src;
x = src.c_str();
return in;
}
friend ostream& operator << (ostream &out, const BigN &x) {
out << x.str();
return out;
}
}A[];
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
std::ios::sync_with_stdio(false);
while (cin >> A[] >> A[]) {
cout << A[] % A[] << endl;
rep(i, ) A[i].clear();
}
return ;
}

hdu 1212 Big Number的更多相关文章

  1. HDU 1212 Big Number(C++ 大数取模)(java 大数类运用)

    Big Number 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1212 ——每天在线,欢迎留言谈论. 题目大意: 给你两个数 n1,n2.其中n1 ...

  2. HDU 1212 Big Number 大数模小数

    http://acm.hdu.edu.cn/showproblem.php?pid=1212 题目大意: 给你一个长度不超过1000的大数A,还有一个不超过100000的B,让你快速求A % B. 什 ...

  3. hdu 1212 Big Number(大数取模)

    Problem Description As we know, Big Number is always troublesome. But it's really important in our A ...

  4. 题解报告:hdu 1212 Big Number(大数取模+同余定理)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1212 Problem Description As we know, Big Number is al ...

  5. hdu 5898 odd-even number 数位DP

    传送门:hdu 5898 odd-even number 思路:数位DP,套着数位DP的模板搞一发就可以了不过要注意前导0的处理,dp[pos][pre][status][ze] pos:当前处理的位 ...

  6. hdu 2665 Kth number

    划分树 /* HDU 2665 Kth number 划分树 */ #include<stdio.h> #include<iostream> #include<strin ...

  7. hdu 4670 Cube number on a tree(点分治)

    Cube number on a tree Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/ ...

  8. 主席树[可持久化线段树](hdu 2665 Kth number、SP 10628 Count on a tree、ZOJ 2112 Dynamic Rankings、codeforces 813E Army Creation、codeforces960F:Pathwalks )

    在今天三黑(恶意评分刷上去的那种)两紫的智推中,突然出现了P3834 [模板]可持久化线段树 1(主席树)就突然有了不详的预感2333 果然...然后我gg了!被大佬虐了! hdu 2665 Kth ...

  9. HDU - 1711 A - Number Sequence(kmp

    HDU - 1711 A - Number Sequence   Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1 ...

随机推荐

  1. 慕课网-安卓工程师初养成-1-1 Java简介

    来源 http://www.imooc.com/video/1430 主要内容 Java平台应用 核心概念:JVM,JDK,JRE 搭建Java开发环境 使用工具开发安卓程序 经验技巧分享 Java历 ...

  2. TCP/IP详解学习笔记(3)-- IP:网际协议

    1.概述      IP是TCP/IP协议族中最为核心的协议.所有的TCP,UDP,ICMP,IGMP数据都以IP数据报格式传输.      IP提供不可靠,无连接的数据报传送服务. 不可靠:它不能保 ...

  3. 浅谈 cookie 和 session

    1.关闭浏览器后,session是否还存在? session在服务器和客户端各保留一个副本,关闭浏览器与否和session是否存在没有任何关系. session采取的是服务器端保持状态的方案,它存储在 ...

  4. C#中DataTable转换JSON

    #region 将DataTable转换为json public string dt2json(DataTable dt) { JavaScriptSerializer jss = new JavaS ...

  5. CSS 三角形绘制方法

    #triangle-up {    width: 0;    height: 0;    border-left: 50px solid transparent;    border-right: 5 ...

  6. 添加favicon.ico网站文件

    <link rel="shortcut icon" type="image/x-icon" href="favicon.ico" me ...

  7. 二十二、OGNL的一些其他操作

    二十二.OGNL的一些其他操作 投影 ?判断满足条件 动作类代码: ^ $   public class Demo2Action extends ActionSupport {     public ...

  8. sql server 查询表信息

    SELECT '表名' = e.[name], '表说明' = f.[value], '字段序号' = a.colorder, '字段名' = a.[name], '字段类型' = b.[name], ...

  9. scan的filter使用

    本次操作的hbase的t1表的数据是: hbase(main)::> scan 't1' ROW COLUMN+CELL column=f1:age, timestamp=, value= co ...

  10. windows下 apache 二级域名相关配置

    小编今天给大家总结下 windows 下 apache的二级域名的相关配置 利用.htaccess将域名绑定到子目录 下面就利用本地127.0.0.1进行测试 我们这里以 www.jobs.com 为 ...