• 大数模板!


      1 #include <iostream>
      2 #include <cstdio>
      3 #include <cstdlib>
      4 #include <cstring>
      5 #define UNIT 10
      6 
      7 using namespace std;
      8 
      9 struct Bignum
     10 {
     11     int val[105];
     12     int len;
     13 
     14     Bignum ()
     15     {
     16         memset (val, 0, sizeof(val));
     17         len = 1;
     18     }
     19 
     20     Bignum operator = (const int &a)
     21     {
     22         int t, p = a;
     23         len = 0;
     24         while (p >= UNIT)
     25         {
     26             t = p - (p/UNIT)*UNIT;
     27             p = p / UNIT;
     28             val[len++] = t;
     29         }
     30         val[len++] = p;
     31         return *this;
     32     }
     33 
     34     Bignum operator + (const Bignum &a) const//大数加大数
     35     {
     36         Bignum x = a;
     37         int L;
     38         L = a.len > len ? a.len : len;
     39         for (int i = 0; i < L; ++i)
     40         {
     41             x.val[i] += val[i];
     42             if (x.val[i] >= UNIT)
     43             {
     44                 x.val[i+1]++;
     45                 x.val[i] -= UNIT;
     46             }
     47         }
     48         if (x.val[L] != 0)
     49             x.len = L+1;
     50         else
     51             x.len = L;
     52         return x;
     53     }
     54 
     55     Bignum operator - (const Bignum &a) const
     56     {
     57         bool flag;
     58         Bignum x1, x2;
     59         if (*this > a)
     60         {
     61             x1 = *this;
     62             x2 = a;
     63             flag = 0;
     64         }
     65         else
     66         {
     67             x1 = a;
     68             x2 = *this;
     69             flag = 1;
     70         }
     71         int j, L = x1.len;
     72         for (int i = 0; i < L; ++i)
     73         {
     74             if (x1.val[i] < x2.val[i])
     75             {
     76                 j = i+1;
     77                 while (x1.val[j] == 0)
     78                     j++;
     79                 x1.val[j--]--;
     80                 while (j > i)
     81                     x1.val[j--] += UNIT-1;
     82                 x1.val[i] += UNIT-x2.val[i];
     83             }
     84             else
     85                 x1.val[i] -= x2.val[i];
     86         }
     87         while (x1.val[x1.len-1] == 0 && x1.len > 1)
     88             x1.len--;
     89         if (flag)
     90             x1.val[x1.len-1] = -x1.val[x1.len-1];
     91         return x1;
     92     }
     93 
     94     Bignum operator * (const Bignum &a) const//大数乘大数
     95     {
     96         Bignum x;
     97         int i, j, up;
     98         int x1, x2;
     99         for (i = 0; i < len; i++)
    100         {
    101             up = 0;
    102             for (j = 0; j < a.len; j++)
    103             {
    104                 x1 = val[i]*a.val[j] + x.val[i+j] + up;
    105                 if (x1 >= UNIT)
    106                 {
    107                     x2 = x1 - x1/UNIT*UNIT;
    108                     up = x1 / UNIT;
    109                     x.val[i+j] = x2;
    110                 }
    111                 else
    112                 {
    113                     up = 0;
    114                     x.val[i+j] = x1;
    115                 }
    116             }
    117             if (up != 0)
    118                 x.val[i+j] = up;
    119         }
    120         x.len = i + j;
    121         while (x.val[x.len-1] == 0 && x.len > 1)
    122             x.len--;
    123         return x;
    124     }
    125 
    126     Bignum operator / (const int &a) const//大数除小数
    127     {
    128         Bignum x;
    129         int down = 0;
    130         for (int i = len-1; i >= 0; --i)
    131         {
    132             x.val[i] = (val[i]+down*UNIT) / a;
    133             down = val[i] + down*UNIT - x.val[i]*a;
    134         }
    135         x.len = len;
    136         while (x.val[x.len-1] == 0 && x.len > 1)
    137             x.len--;
    138         return x;
    139     }
    140 
    141     int operator % (const int &a) const//大数模小数
    142     {
    143         int x = 0;
    144         for (int i = len-1; i >= 0; --i)
    145             x = ((x*UNIT)%a+val[i]) % a;
    146         return x;
    147     }
    148 
    149     Bignum operator ^ (const int &a) const
    150     {
    151         Bignum p, x;
    152         x.val[0] = 1;
    153         if(a < 0)
    154             exit(-1);
    155         if(a == 0)
    156             return x;
    157         if(a == 1)
    158             return *this;
    159         int n = a, i;
    160         while(n > 1)
    161         {
    162             p = *this;
    163             for(i = 1; (i<<1) <= n; i<<=1)
    164                 p = p * p;
    165             n -= i;
    166             x = x * p;
    167             if(n == 1)
    168                 x = x * (*this);
    169         }
    170         return x;
    171     }
    172 
    173     bool operator > (const Bignum &a) const
    174     {
    175         int now;
    176         if (len > a.len)
    177             return true;
    178         else if (len == a.len)
    179         {
    180             now = len - 1;
    181             while (val[now] == a.val[now] && now >= 0)
    182                 now--;
    183             if(now >= 0 && val[now] > a.val[now])
    184                 return true;
    185             else
    186                 return false;
    187         }
    188         else
    189             return false;
    190     }
    191 };
    192 
    193 int main()
    194 {
    195     /*f[0] = 1;
    196     for(int i=1;i<=100;i++)
    197     f[i]=f[i-1]*(4*i-2)/(i+1);//卡特兰数递推式
    198     int n;
    199     while(scanf("%d",&n)==1)
    200     {
    201         if(n==-1)
    202             break;
    203         f[n].print();
    204     }*/
    205     return 0;
    206 }
    View Code
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  • 原文地址:https://www.cnblogs.com/M-D-LUFFI/p/4042461.html
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