算法提高 P1001

      必须感叹下，大数模板就是好用！

    AC代码：

#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <utility>#include <string>#include <iostream>#include <map>#include <set>#include <vector>#include <queue>#include <stack>using namespace std;#define eps 1e-10#define inf 0x3f3f3f3f#define PI pair<int, int> typedef long long LL;const int maxn = 1e4 + 5;struct BigInteger {vector<int>s;  //12345--54321void DealZero() { //处理前导0 for(int i = s.size() - 1; i > 0; --i){if(s[i] == 0) s.pop_back();else break;}} BigInteger operator = (long long num) { // 赋值运算符 s.clear(); vector<int>tmp;       do{       s.push_back(num % 10);       num /= 10;}while(num);return *this;    }    BigInteger operator = (const string& str) { // 赋值运算符    s.clear();   for(int i = str.size() - 1; i >= 0; --i) s.push_back(str[i] - '0');   this->DealZero();   return *this;    }        BigInteger operator + (const BigInteger& b) const {    BigInteger c;    c.s.clear();    int len1 = s.size(), len2 = b.s.size();     for(int i = 0, g = 0; g > 0 || i < len1 || i < len2; ++i) {    int x = g;    if(i < len1) x += s[i];    if(i < len2) x += b.s[i];    c.s.push_back(x % 10);    g = x / 10;}    return c;   }      //大数减小数    BigInteger operator - (const BigInteger& b) const {    BigInteger c;    c.s.clear();    int len1 = s.size(), len2 = b.s.size();     for(int i = 0, g = 0; i < len1 || i < len2; ++i) {    int x = g;    if(i < len1) x += s[i];    g = 0;    if(i < len2) x -= b.s[i];    if(x < 0) {    g = -1; //借位     x += 10;}     c.s.push_back(x);}c.DealZero();    return c;   }      BigInteger operator * (const BigInteger& b) const {   BigInteger c, tmp;    c.s.clear();    int len1 = s.size(), len2 = b.s.size();   for(int i = 0; i < len1; ++i) {   tmp.s.clear();tmp;   int num = i;while(num--) tmp.s.push_back(0);int g = 0;for(int j = 0; j < len2; ++j) {int x = s[i] * b.s[j] + g;tmp.s.push_back(x % 10);g = x / 10;}if(g > 0) tmp.s.push_back(g);c = c + tmp;}c.DealZero();return c;   }      //单精度除法    BigInteger operator / (const int b) const {    BigInteger c, tmp;   c.s.clear();   int len = s.size();   int div = 0;   for(int i = len - 1; i >= 0; --i) {   div = div * 10 + s[i];   while(div < b && i > 0) {   div = div * 10 + s[--i];}tmp.s.push_back(div / b);div %= b;}for(int i = tmp.s.size() - 1; i >= 0; --i) c.s.push_back(tmp.s[i]);c.DealZero();return c;   }       bool operator < (const BigInteger& b) const {      int len1 = s.size(), len2 = b.s.size();      if(len1 != len2) return len1 < len2;      for(int i = len1 - 1; i >= 0; --i) {      if(s[i] != b.s[i]) return s[i] < b.s[i];}return false; //相等    }      bool operator <= (const BigInteger& b) const {   return !(b < *this);   }   string ToStr() {   string ans;   ans.clear();   for(int i = s.size()-1; i >= 0; --i)    ans.push_back(s[i] + '0');   return ans;   }      //大数开方 /**大数开方用法说明：   字符串必须从第二个位置开始输入,且s[0] = '0'    scanf("%s", s+1); */    BigInteger SQRT(char *s) {   string p = "";s[0]='0';    if(strlen(s)%2 == 1)        work(p, 2, s+1, 0);    else        work(p, 2, s, 0);   BigInteger c;   c.s.clear();   c = p;   return c;   }      //开方准备  //------------------------------------int l;int work(string &p, int o,char *O,int I){    char c, *D=O ;    if(o>0)    {        for(l=0;D[l];D[l++]-=10)        {            D[l++]-=120;            D[l]-=110;            while(!work(p, 0, O, l))                D[l]+=20;            p += (char)((D[l]+1032)/20);                    }    }    else    {        c=o+(D[I]+82)%10-(I>l/2)*(D[I-l+I]+72)/10-9;        D[I]+=I<0 ? 0 : !(o=work(p, c/10,O,I-1))*((c+999)%10-(D[I]+92)%10);    }    return o;}//-----------------------------------------};ostream& operator << (ostream &out, const BigInteger& x) {for(int i = x.s.size() - 1; i >= 0; --i)out << x.s[i];return out;}istream& operator >> (istream &in, BigInteger& x) {  string s;  if(!(in >> s)) return in;  x = s;  return in;}int main() {BigInteger a, b;while(cin >> a >> b) {cout << a * b << "\n";}return 0;}

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