UVa：10494 - If We Were a Child Again

刘汝佳的《算法竞赛入门经典》第五章中有高精度运算类 bign，如下：

#include<cstdio>
#include<iostream>
using namespace std;

const int maxn = 200;
struct bign{
  int len, s[maxn];

  bign() {
    memset(s, 0, sizeof(s));
    len = 1;
  }

  bign(int num) {
    *this = num;
  }

  bign(const char* num) {
    *this = num;
  }

  bign operator = (int num) {
    char s[maxn];
    sprintf(s, "%d", num);
    *this = s;
    return *this;
  }

  bign operator = (const char* num) {
    len = strlen(num);
    for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
    return *this;
  }

  string str() const {
    string res = "";
    for(int i = 0; i < len; i++) res = (char)(s[i] + '0') + res;
    if(res == "") res = "0";
    return res;
  }
//高精度加法
  bign operator + (const bign& b) const{
    bign c;
    c.len = 0;
    for(int i = 0, g = 0; g || i < max(len, b.len); i++) {
      int x = g;
      if(i < len) x += s[i];
      if(i < b.len) x += b.s[i];
      c.s[c.len++] = x % 10;
      g = x / 10;
    }
    return c;
  }
//忽略前导0
  void clean() {
    while(len > 1 && !s[len-1]) len--;
  }
//高精度乘法
  bign operator * (const bign& b) {
    bign c; c.len = len + b.len;
    for(int i = 0; i < len; i++)
      for(int j = 0; j < b.len; j++)
        c.s[i+j] += s[i] * b.s[j];
    for(int i = 0; i < c.len-1; i++){
      c.s[i+1] += c.s[i] / 10;
      c.s[i] %= 10;
    }
    c.clean();
    return c;
  }
//高精度减法
  bign operator - (const bign& b) {
    bign c; c.len = 0;
    for(int i = 0, g = 0; i < len; i++) {
      int x = s[i] - g;
      if(i < b.len) x -= b.s[i];
      if(x >= 0) g = 0;
      else {
        g = 1;
        x += 10;
      }
      c.s[c.len++] = x;
    }
    c.clean();
    return c;
  }
 //高精度除以低精度
  bign operator / (int b) const{
        assert(b > 0);
        bign c;c.len = len;
        for (int i = len-1, g = 0; i >= 0; --i){
            int x = 10*g+s[i];
            c.s[i] = x/b;
            g = x-c.s[i]*b;
        }
        c.clean();
        return c;
  }
//高精度对低精度取余
  bign operator % (int b){
        assert(b > 0);
        bign d = b;
        bign c = *this-*this/b*d;
        return c;
  }

  bool operator < (const bign& b) const{
    if(len != b.len) return len < b.len;
    for(int i = len-1; i >= 0; i--)
      if(s[i] != b.s[i]) return s[i] < b.s[i];
    return false;
  }

  bool operator > (const bign& b) const{
    return b < *this;
  }

  bool operator <= (const bign& b) {
    return !(b > *this);
  }

  bool operator == (const bign& b) {
    return !(b < *this) && !(*this < b);
  }

  bign operator += (const bign& b) {
    *this = *this + b;
    return *this;
  }
};

istream& operator >> (istream &in, bign& x) {
  string s;
  in >> s;
  x = s.c_str();
  return in;
}

ostream& operator << (ostream &out, const bign& x) {
  out << x.str();
  return out;
}

int main() {
  bign a;
  cin >> a;
  a += "123456789123456789000000000";
  cout << a*2 << endl;
  return 0;
}

将上诉模板稍加修改，即可解决此题。
特别要注意的是：题目所描述的第一个数据有可能大于int的最大值231-1，所以用long long定义整型数据，预防溢出。
AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<climits>
#include<cassert>
using namespace std;
const int maxn=30000;

struct bign{
  long long len, s[maxn];

  bign() {
    memset(s, 0, sizeof(s));
    len = 1;
  }

  bign(long long num) {
    *this = num;
  }

  bign(const char* num) {
    *this = num;
  }

  bign operator = (long long num) {
    char s[maxn];
    sprintf(s, "%d", num);
    *this = s;
    return *this;
  }

  bign operator = (const char* num) {
    len = strlen(num);
    for(long long i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
    return *this;
  }

  string str() const {
    string res = "";
    for(long long i = 0; i < len; i++) res = (char)(s[i] + '0') + res;
    if(res == "") res = "0";
    return res;
  }

  bign operator + (const bign& b) const{
    bign c;
    c.len = 0;
    for(long long i = 0, g = 0; g || i < max(len, b.len); i++) {
      long long x = g;
      if(i < len) x += s[i];
      if(i < b.len) x += b.s[i];
      c.s[c.len++] = x % 10;
      g = x / 10;
    }
    return c;
  }

  void clean() {
    while(len > 1 && !s[len-1]) len--;
  }

  bign operator * (const bign& b) {
    bign c; c.len = len + b.len;
    for(long long i = 0; i < len; i++)
      for(long long j = 0; j < b.len; j++)
        c.s[i+j] += s[i] * b.s[j];
    for(long long i = 0; i < c.len-1; i++){
      c.s[i+1] += c.s[i] / 10;
      c.s[i] %= 10;
    }
    c.clean();
    return c;
  }

  bign operator - (const bign& b) {
    bign c; c.len = 0;
    for(long long i = 0, g = 0; i < len; i++) {
      long long x = s[i] - g;
      if(i < b.len) x -= b.s[i];
      if(x >= 0) g = 0;
      else {
        g = 1;
        x += 10;
      }
      c.s[c.len++] = x;
    }
    c.clean();
    return c;
  }

  bign operator / (long long b) const{
        assert(b > 0);
        bign c;c.len = len;
        for (long long i = len-1, g = 0; i >= 0; --i){
            long long x = 10*g+s[i];
            c.s[i] = x/b;
            g = x-c.s[i]*b;
        }
        c.clean();
        return c;
  }

  bign operator % (long long b){
        assert(b > 0);
        bign d = b;
        bign c = *this-*this/b*d;
        return c;
  }

  bool operator < (const bign& b) const{
    if(len != b.len) return len < b.len;
    for(long long i = len-1; i >= 0; i--)
      if(s[i] != b.s[i]) return s[i] < b.s[i];
    return false;
  }

  bool operator > (const bign& b) const{
    return b < *this;
  }

  bool operator <= (const bign& b) {
    return !(b > *this);
  }

  bool operator == (const bign& b) {
    return !(b < *this) && !(*this < b);
  }

  bign operator += (const bign& b) {
    *this = *this + b;
    return *this;
  }
};

istream& operator >> (istream &in, bign& x) {
  string s;
  in >> s;
  x = s.c_str();
  return in;
}

ostream& operator << (ostream &out, const bign& x) {
  out << x.str();
  return out;
}


int main()
{
    bign a;
    char b;
    long long c;
    while (cin>>a>>b>>c){
        if (b=='/')
            cout<<a/c<<endl;
        else
            cout<<a%c<<endl;
    }
    return 0;
}

还可以参考这个AC代码，很是精辟哦（没有用到bign类）： 点击打开链接

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