• hdu 1212 Big Number


    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。。

      1 #include<algorithm>
      2 #include<iostream>
      3 #include<cstdlib>
      4 #include<cstring>
      5 #include<cassert>
      6 #include<cstdio>
      7 #include<vector>
      8 #include<string>
      9 #include<map>
     10 #include<set>
     11 using std::cin;
     12 using std::max;
     13 using std::cout;
     14 using std::endl;
     15 using std::string;
     16 using std::istream;
     17 using std::ostream;
     18 #define sz(c) (int)(c).size()
     19 #define all(c) (c).begin(), (c).end()
     20 #define iter(c) decltype((c).begin())
     21 #define cls(arr,val) memset(arr,val,sizeof(arr))
     22 #define cpresent(c, e) (find(all(c), (e)) != (c).end())
     23 #define rep(i, n) for (int i = 0; i < (int)(n); i++)
     24 #define fork(i, k, n) for (int i = (int)k; i <= (int)n; i++)
     25 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)
     26 #define pb(e) push_back(e)
     27 #define mp(a, b) make_pair(a, b)
     28 struct BigN {
     29     typedef unsigned long long ull;
     30     static const int Max_N = 2010;
     31     int len, data[Max_N];
     32     BigN() { memset(data, 0, sizeof(data)), len = 0; }
     33     BigN(const int num) {
     34         memset(data, 0, sizeof(data));
     35         *this = num;
     36     }
     37     BigN(const char *num) {
     38         memset(data, 0, sizeof(data));
     39         *this = num;
     40     }
     41     void clear() { len = 0, memset(data, 0, sizeof(data)); }
     42     BigN& clean(){ while (len > 1 && !data[len - 1]) len--;  return *this; }
     43     string str() const {
     44         string res = "";
     45         for (int i = len - 1; ~i; i--) res += (char)(data[i] + '0');
     46         if (res == "") res = "0";
     47         res.reserve();
     48         return res;
     49     }
     50     BigN operator = (const int num) {
     51         int j = 0, i = num;
     52         do data[j++] = i % 10; while (i /= 10);
     53         len = j;
     54         return *this;
     55     }
     56     BigN operator = (const char *num) {
     57         len = strlen(num);
     58         for (int i = 0; i < len; i++) data[i] = num[len - i - 1] - '0';
     59         return *this;
     60     }
     61     BigN operator + (const BigN &x) const {
     62         BigN res;
     63         int n = max(len, x.len) + 1;
     64         for (int i = 0, g = 0; i < n; i++) {
     65             int c = data[i] + x.data[i] + g;
     66             res.data[res.len++] = c % 10;
     67             g = c / 10;
     68         }
     69         return res.clean();
     70     }
     71     BigN operator * (const BigN &x) const {
     72         BigN res;
     73         int n = x.len;
     74         res.len = n + len;
     75         for (int i = 0; i < len; i++) {
     76             for (int j = 0, g = 0; j < n; j++) {
     77                 res.data[i + j] += data[i] * x.data[j];
     78             }
     79         }
     80         for (int i = 0; i < res.len - 1; i++) {
     81             res.data[i + 1] += res.data[i] / 10;
     82             res.data[i] %= 10;
     83         }
     84         return res.clean();
     85     }
     86     BigN operator * (const int num) const {
     87         BigN res;
     88         res.len = len + 1;
     89         for (int i = 0, g = 0; i < len; i++) res.data[i] *= num;
     90         for (int i = 0; i < res.len - 1; i++) {
     91             res.data[i + 1] += res.data[i] / 10;
     92             res.data[i] %= 10;
     93         }
     94         return res.clean();
     95     }
     96     BigN operator - (const BigN &x) const {
     97         assert(x <= *this);
     98         BigN res;
     99         for (int i = 0, g = 0; i < len; i++) {
    100             int c = data[i] - g;
    101             if (i < x.len) c -= x.data[i];
    102             if (c >= 0) g = 0;
    103             else g = 1, c += 10;
    104             res.data[res.len++] = c;
    105         }
    106         return res.clean();
    107     }
    108     BigN operator / (const BigN &x) const {
    109         BigN res, f = 0;
    110         for (int i = len - 1; ~i; i--) {
    111             f *= 10;
    112             f.data[0] = data[i];
    113             while (f >= x) {
    114                 f -= x;
    115                 res.data[i]++;
    116             }
    117         }
    118         res.len = len;
    119         return res.clean();
    120     }
    121     BigN operator % (const BigN &x) {
    122         BigN res = *this / x;
    123         res = *this - res * x;
    124         return res;
    125     }
    126     BigN operator += (const BigN &x) { return *this = *this + x; }
    127     BigN operator *= (const BigN &x) { return *this = *this * x; }
    128     BigN operator -= (const BigN &x) { return *this = *this - x; }
    129     BigN operator /= (const BigN &x) { return *this = *this / x; }
    130     BigN operator %= (const BigN &x) { return *this = *this % x; }
    131     bool operator <  (const BigN &x) const {
    132         if (len != x.len) return len < x.len;
    133         for (int i = len - 1; ~i; i--) {
    134             if (data[i] != x.data[i]) return data[i] < x.data[i];
    135         }
    136         return false;
    137     }
    138     bool operator >(const BigN &x) const { return x < *this; }
    139     bool operator<=(const BigN &x) const { return !(x < *this); }
    140     bool operator>=(const BigN &x) const { return !(*this < x); }
    141     bool operator!=(const BigN &x) const { return x < *this || *this < x; }
    142     bool operator==(const BigN &x) const { return !(x < *this) && !(x > *this); }
    143     friend istream& operator >> (istream &in, BigN &x) {
    144         string src;
    145         in >> src;
    146         x = src.c_str();
    147         return in;
    148     }
    149     friend ostream& operator << (ostream &out, const BigN &x) {
    150         out << x.str();
    151         return out;
    152     }
    153 }A[2];
    154 int main() {
    155 #ifdef LOCAL
    156     freopen("in.txt", "r", stdin);
    157     freopen("out.txt", "w+", stdout);
    158 #endif
    159     std::ios::sync_with_stdio(false);
    160     while (cin >> A[0] >> A[1]) {
    161         cout << A[0] % A[1] << endl;
    162         rep(i, 2) A[i].clear();
    163     }
    164     return 0;
    165 }
    View Code
    By: GadyPu 博客地址:http://www.cnblogs.com/GadyPu/ 转载请说明
  • 相关阅读:
    CPU 上下文切换是什么
    Linux性能优化实战
    JavaScript 概述
    最全前端资源汇总
    zabbix 源码安装
    单例
    php防止sql注入
    python 多进程读写文件
    python twisted异步将数据导入到数据库中
    scrapy-splash常用设置
  • 原文地址:https://www.cnblogs.com/GadyPu/p/4575401.html
Copyright © 2020-2023  润新知