• 高精度


    是时候结束这个局面了。。

    说真的,有了模板一切都好说。

    A+B

      1 #include <cstdio>
      2 #include <cstring>
      3 #include <iostream>
      4 
      5 using namespace std;
      6 
      7 const int MAXN = 10001;
      8 
      9 char c[MAXN];
     10 
     11 inline char *read()
     12 {
     13     scanf("%s", c);
     14     return c;
     15 }
     16 
     17 struct Big_int
     18 {
     19     int s[MAXN], idx;
     20     Big_int()
     21     {
     22         idx = 0;
     23         memset(s, 0, sizeof(s));
     24     }
     25     inline void operator = (char *c)
     26     {
     27         idx = strlen(c);
     28         for(int i = 0; i < idx; i++) s[i] = c[idx - i - 1] - '0';
     29     }
     30     inline void operator = (int x)
     31     {
     32         while(x)
     33         {
     34             s[idx] = x % 10;
     35             x /= 10;
     36             idx++;    
     37         }
     38     }
     39     inline void print()
     40     {
     41         if(!idx) printf("0");
     42         else for(int i = idx - 1; i >= 0; i--) printf("%d", s[i]);
     43         puts("");
     44     }
     45 };
     46 
     47 inline Big_int operator + (const Big_int x, const Big_int y)
     48 {
     49     Big_int ret;
     50     ret.idx = max(x.idx, y.idx) + 1;
     51     for(int i = 0; i < ret.idx; i++)
     52     {
     53         ret.s[i] += x.s[i] + y.s[i];
     54         if(ret.s[i] >= 10)
     55             ret.s[i + 1] += 1, ret.s[i] -= 10;
     56     }
     57     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
     58     return ret;
     59 }
     60 
     61 inline bool operator < (const Big_int x, const Big_int y)
     62 {
     63     if(x.idx < y.idx) return 1;
     64     if(x.idx > y.idx) return 0;
     65     for(int i = x.idx - 1; i >= 0; i--)
     66         if(x.s[i] ^ y.s[i])
     67             return x.s[i] < y.s[i];
     68     return 0;
     69 }
     70 
     71 inline Big_int operator - (Big_int x, Big_int y)
     72 {
     73     Big_int ret;
     74     if(x < y) swap(x, y);
     75     ret.idx = x.idx;
     76     for(int i = 0; i < ret.idx; i++)
     77     {
     78         if(x.s[i] < y.s[i])
     79         {
     80             x.s[i] += 10;
     81             x.s[i + 1]--;
     82         }
     83         ret.s[i] = x.s[i] - y.s[i];
     84     }
     85     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
     86     return ret;
     87 }
     88 
     89 inline Big_int operator * (const Big_int x, const Big_int y)
     90 {
     91     Big_int ret;
     92     ret.idx = x.idx + y.idx;
     93     for(int i = 0; i < x.idx; i++)
     94         for(int j = 0; j < y.idx; j++)
     95         {
     96             ret.s[i + j] += x.s[i] * y.s[j];
     97             ret.s[i + j + 1] += ret.s[i + j] / 10;
     98             ret.s[i + j] %= 10;
     99         }
    100     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
    101     return ret;
    102 }
    103 
    104 Big_int a, b, ans;
    105 
    106 int main()
    107 {
    108     a = read();
    109     b = read();
    110     ans = a + b;
    111     ans.print();
    112     return 0;
    113 }
    View Code

    A-B

     1 #include <cstdio>
     2 #include <cstring>
     3 #include <iostream>
     4 
     5 using namespace std;
     6 
     7 const int MAXN = 10001;
     8 
     9 char c[MAXN];
    10 
    11 inline char *read()
    12 {
    13     scanf("%s", c);
    14     return c;
    15 }
    16 
    17 struct Big_int
    18 {
    19     int s[MAXN], idx;
    20     Big_int()
    21     {
    22         idx = 0;
    23         memset(s, 0, sizeof(s));
    24     }
    25     inline void operator = (char *c)
    26     {
    27         idx = strlen(c);
    28         for(int i = 0; i < idx; i++) s[i] = c[idx - i - 1] - '0';
    29     }
    30     inline void operator = (int x)
    31     {
    32         while(x)
    33         {
    34             s[idx] = x % 10;
    35             x /= 10;
    36             idx++;    
    37         }
    38     }
    39     inline void print()
    40     {
    41         if(!idx) printf("0");
    42         else for(int i = idx - 1; i >= 0; i--) printf("%d", s[i]);
    43         puts("");
    44     }
    45 };
    46 
    47 inline Big_int operator + (const Big_int x, const Big_int y)
    48 {
    49     Big_int ret;
    50     ret.idx = max(x.idx, y.idx) + 1;
    51     for(int i = 0; i < ret.idx; i++)
    52     {
    53         ret.s[i] += x.s[i] + y.s[i];
    54         ret.s[i + 1] += ret.s[i] / 10;
    55         ret.s[i] %= 10;
    56     }
    57     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
    58     return ret;
    59 }
    60 
    61 inline bool operator < (const Big_int x, const Big_int y)
    62 {
    63     if(x.idx < y.idx) return 1;
    64     if(x.idx > y.idx) return 0;
    65     for(int i = x.idx - 1; i >= 0; i--)
    66         if(x.s[i] ^ y.s[i])
    67             return x.s[i] < y.s[i];
    68     return 0;
    69 }
    70 
    71 inline Big_int operator - (Big_int x, Big_int y)
    72 {
    73     Big_int ret;
    74     if(x < y) swap(x, y);
    75     ret.idx = x.idx;
    76     for(int i = 0; i < ret.idx; i++)
    77     {
    78         if(x.s[i] < y.s[i])
    79         {
    80             x.s[i] += 10;
    81             x.s[i + 1]--;
    82         }
    83         ret.s[i] = x.s[i] - y.s[i];
    84     }
    85     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
    86     return ret;
    87 }
    88 
    89 Big_int a, b, ans;
    90 
    91 int main()
    92 {
    93     a = read();
    94     b = read();
    95     if(a < b) printf("-");
    96     ans = a - b;
    97     ans.print();
    98     return 0;
    99 }
    View Code

    A*B

      1 #include <cstdio>
      2 #include <cstring>
      3 #include <iostream>
      4 
      5 using namespace std;
      6 
      7 const int MAXN = 10001;
      8 
      9 char c[MAXN];
     10 
     11 inline char *read()
     12 {
     13     scanf("%s", c);
     14     return c;
     15 }
     16 
     17 struct Big_int
     18 {
     19     int s[MAXN], idx;
     20     Big_int()
     21     {
     22         idx = 0;
     23         memset(s, 0, sizeof(s));
     24     }
     25     inline void operator = (char *c)
     26     {
     27         idx = strlen(c);
     28         for(int i = 0; i < idx; i++) s[i] = c[idx - i - 1] - '0';
     29     }
     30     inline void operator = (int x)
     31     {
     32         while(x)
     33         {
     34             s[idx] = x % 10;
     35             x /= 10;
     36             idx++;    
     37         }
     38     }
     39     inline void print()
     40     {
     41         if(!idx) printf("0");
     42         else for(int i = idx - 1; i >= 0; i--) printf("%d", s[i]);
     43         puts("");
     44     }
     45 };
     46 
     47 inline Big_int operator + (const Big_int x, const Big_int y)
     48 {
     49     Big_int ret;
     50     ret.idx = max(x.idx, y.idx) + 1;
     51     for(int i = 0; i < ret.idx; i++)
     52     {
     53         ret.s[i] += x.s[i] + y.s[i];
     54         if(ret.s[i] >= 10)
     55             ret.s[i + 1] += 1, ret.s[i] -= 10;
     56     }
     57     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
     58     return ret;
     59 }
     60 
     61 inline bool operator < (const Big_int x, const Big_int y)
     62 {
     63     if(x.idx < y.idx) return 1;
     64     if(x.idx > y.idx) return 0;
     65     for(int i = x.idx - 1; i >= 0; i--)
     66         if(x.s[i] ^ y.s[i])
     67             return x.s[i] < y.s[i];
     68     return 0;
     69 }
     70 
     71 inline Big_int operator - (Big_int x, Big_int y)
     72 {
     73     Big_int ret;
     74     if(x < y) swap(x, y);
     75     ret.idx = x.idx;
     76     for(int i = 0; i < ret.idx; i++)
     77     {
     78         if(x.s[i] < y.s[i])
     79         {
     80             x.s[i] += 10;
     81             x.s[i + 1]--;
     82         }
     83         ret.s[i] = x.s[i] - y.s[i];
     84     }
     85     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
     86     return ret;
     87 }
     88 
     89 inline Big_int operator * (const Big_int x, const Big_int y)
     90 {
     91     Big_int ret;
     92     ret.idx = x.idx + y.idx;
     93     for(int i = 0; i < x.idx; i++)
     94         for(int j = 0; j < y.idx; j++)
     95         {
     96             ret.s[i + j] += x.s[i] * y.s[j];
     97             ret.s[i + j + 1] += ret.s[i + j] / 10;
     98             ret.s[i + j] %= 10;
     99         }
    100     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
    101     return ret;
    102 }
    103 
    104 Big_int a, b, ans;
    105 
    106 int main()
    107 {
    108     a = read();
    109     b = read();
    110     ans = a * b;
    111     ans.print();
    112     return 0;
    113 }
    View Code

    数楼梯(高精fib)

      1 #include <cstdio>
      2 #include <cstring>
      3 #include <iostream>
      4 
      5 using namespace std;
      6 
      7 const int MAXN = 10001;
      8 
      9 char c[MAXN];
     10 
     11 inline char *read()
     12 {
     13     scanf("%s", c);
     14     return c;
     15 }
     16 
     17 struct Big_int
     18 {
     19     int s[MAXN], idx;
     20     Big_int()
     21     {
     22         idx = 0;
     23         memset(s, 0, sizeof(s));
     24     }
     25     inline void operator = (char *c)
     26     {
     27         idx = strlen(c);
     28         for(int i = 0; i < idx; i++) s[i] = c[idx - i - 1] - '0';
     29     }
     30     inline void operator = (int x)
     31     {
     32         while(x)
     33         {
     34             s[idx] = x % 10;
     35             x /= 10;
     36             idx++;    
     37         }
     38     }
     39     inline void print()
     40     {
     41         if(!idx) printf("0");
     42         else for(int i = idx - 1; i >= 0; i--) printf("%d", s[i]);
     43         puts("");
     44     }
     45 };
     46 
     47 inline Big_int operator + (const Big_int x, const Big_int y)
     48 {
     49     Big_int ret;
     50     ret.idx = max(x.idx, y.idx) + 1;
     51     for(int i = 0; i < ret.idx; i++)
     52     {
     53         ret.s[i] += x.s[i] + y.s[i];
     54         if(ret.s[i] >= 10)
     55             ret.s[i + 1] += 1, ret.s[i] -= 10;
     56     }
     57     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
     58     return ret;
     59 }
     60 
     61 inline bool operator < (const Big_int x, const Big_int y)
     62 {
     63     if(x.idx < y.idx) return 1;
     64     if(x.idx > y.idx) return 0;
     65     for(int i = x.idx - 1; i >= 0; i--)
     66         if(x.s[i] ^ y.s[i])
     67             return x.s[i] < y.s[i];
     68     return 0;
     69 }
     70 
     71 inline Big_int operator - (Big_int x, Big_int y)
     72 {
     73     Big_int ret;
     74     if(x < y) swap(x, y);
     75     ret.idx = x.idx;
     76     for(int i = 0; i < ret.idx; i++)
     77     {
     78         if(x.s[i] < y.s[i])
     79         {
     80             x.s[i] += 10;
     81             x.s[i + 1]--;
     82         }
     83         ret.s[i] = x.s[i] - y.s[i];
     84     }
     85     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
     86     return ret;
     87 }
     88 
     89 inline Big_int operator * (const Big_int x, const Big_int y)
     90 {
     91     Big_int ret;
     92     ret.idx = x.idx + y.idx;
     93     for(int i = 0; i < x.idx; i++)
     94         for(int j = 0; j < y.idx; j++)
     95         {
     96             ret.s[i + j] += x.s[i] * y.s[j];
     97             ret.s[i + j + 1] += ret.s[i + j] / 10;
     98             ret.s[i + j] %= 10;
     99         }
    100     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
    101     return ret;
    102 }
    103 
    104 int n;
    105 Big_int a, b, ans;
    106 
    107 int main()
    108 {
    109     scanf("%d", &n);
    110     a = 0;
    111     b = 1;
    112     for(int i = 1; i <= n; i++)
    113     {
    114         ans = a + b;
    115         a = b;
    116         b = ans;
    117     }
    118     ans.print();
    119     return 0;
    120 }
    View Code

    B进制星球(随便改改模板即可)

      1 #include <cstdio>
      2 #include <cstring>
      3 #include <iostream>
      4 
      5 using namespace std;
      6 
      7 const int MAXN = 10001;
      8 
      9 int n;
     10 char c[MAXN];
     11 
     12 inline char *read()
     13 {
     14     scanf("%s", c);
     15     return c;
     16 }
     17 
     18 struct Big_int
     19 {
     20     int s[MAXN], idx;
     21     Big_int()
     22     {
     23         idx = 0;
     24         memset(s, 0, sizeof(s));
     25     }
     26     inline void operator = (char *c)
     27     {
     28         idx = strlen(c);
     29         for(int i = 0; i < idx; i++)
     30             if(isdigit(c[idx - i - 1])) s[i] = c[idx - i - 1] - '0';
     31             else s[i] = c[idx - i - 1] - 'A' + 10;
     32     }
     33     inline void operator = (int x)
     34     {
     35         while(x)
     36         {
     37             s[idx] = x % 10;
     38             x /= 10;
     39             idx++;
     40         }
     41     }
     42     inline void print()
     43     {
     44         if(!idx) printf("0");
     45         else for(int i = idx - 1; i >= 0; i--)
     46             if(s[i] >= 10) printf("%c", char(s[i] - 10 + 'A'));
     47             else printf("%d", s[i]);
     48         puts("");
     49     }
     50 };
     51 
     52 inline Big_int operator + (const Big_int x, const Big_int y)
     53 {
     54     Big_int ret;
     55     ret.idx = max(x.idx, y.idx) + 1;
     56     for(int i = 0; i < ret.idx; i++)
     57     {
     58         ret.s[i] += x.s[i] + y.s[i];
     59         if(ret.s[i] >= n)
     60             ret.s[i] -= n, ret.s[i + 1] += 1;
     61     }
     62     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
     63     return ret;
     64 }
     65 
     66 inline bool operator < (const Big_int x, const Big_int y)
     67 {
     68     if(x.idx < y.idx) return 1;
     69     if(x.idx > y.idx) return 0;
     70     for(int i = x.idx - 1; i >= 0; i--)
     71         if(x.s[i] ^ y.s[i])
     72             return x.s[i] < y.s[i];
     73     return 0;
     74 }
     75 
     76 inline Big_int operator - (Big_int x, Big_int y)
     77 {
     78     Big_int ret;
     79     if(x < y) swap(x, y);
     80     ret.idx = x.idx;
     81     for(int i = 0; i < ret.idx; i++)
     82     {
     83         if(x.s[i] < y.s[i])
     84         {
     85             x.s[i] += 10;
     86             x.s[i + 1]--;
     87         }
     88         ret.s[i] = x.s[i] - y.s[i];
     89     }
     90     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
     91     return ret;
     92 }
     93 
     94 inline Big_int operator * (const Big_int x, const Big_int y)
     95 {
     96     Big_int ret;
     97     ret.idx = x.idx + y.idx;
     98     for(int i = 0; i < x.idx; i++)
     99         for(int j = 0; j < y.idx; j++)
    100         {
    101             ret.s[i + j] += x.s[i] * y.s[j];
    102             ret.s[i + j + 1] += ret.s[i + j] / 10;
    103             ret.s[i + j] %= 10;
    104         }
    105     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
    106     return ret;
    107 }
    108 
    109 Big_int a, b, ans;
    110 
    111 int main()
    112 {
    113     scanf("%d", &n);
    114     a = read();
    115     b = read();
    116     ans = a + b;
    117     ans.print();
    118     return 0;
    119 }
    View Code

    国王游戏。。。

    (高精除以低精,更好写。。)

      1 #include <cstdio>
      2 #include <cstring>
      3 #include <iostream>
      4 #include <algorithm>
      5 #define N 1001
      6 
      7 using namespace std;
      8 
      9 int n, b;
     10 
     11 struct node
     12 {
     13     int a, b;
     14 }p[N];
     15 
     16 const int MAXN = 10001;
     17 
     18 char c[MAXN];
     19 
     20 inline char *read()
     21 {
     22     scanf("%s", c);
     23     return c;
     24 }
     25 
     26 struct Big_int
     27 {
     28     int s[MAXN], idx;
     29     Big_int()
     30     {
     31         idx = 0;
     32         memset(s, 0, sizeof(s));
     33     }
     34     inline void operator = (char *c)
     35     {
     36         idx = strlen(c);
     37         for(int i = 0; i < idx; i++) s[i] = c[idx - i - 1] - '0';
     38     }
     39     inline void operator = (int x)
     40     {
     41         idx = 0;
     42         while(x)
     43         {
     44             s[idx] = x % 10;
     45             x /= 10;
     46             idx++;    
     47         }
     48     }
     49     inline void print()
     50     {
     51         if(!idx) printf("0");
     52         else for(int i = idx - 1; i >= 0; i--) printf("%d", s[i]);
     53         puts("");
     54     }
     55 };
     56 
     57 inline Big_int operator + (const Big_int x, const Big_int y)
     58 {
     59     Big_int ret;
     60     ret.idx = max(x.idx, y.idx) + 1;
     61     for(int i = 0; i < ret.idx; i++)
     62     {
     63         ret.s[i] += x.s[i] + y.s[i];
     64         if(ret.s[i] >= 10)
     65             ret.s[i + 1] += 1, ret.s[i] -= 10;
     66     }
     67     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
     68     return ret;
     69 }
     70 
     71 inline bool operator < (const Big_int x, const Big_int y)
     72 {
     73     if(x.idx < y.idx) return 1;
     74     if(x.idx > y.idx) return 0;
     75     for(int i = x.idx - 1; i >= 0; i--)
     76         if(x.s[i] ^ y.s[i])
     77             return x.s[i] < y.s[i];
     78     return 0;
     79 }
     80 
     81 inline Big_int operator - (Big_int x, Big_int y)
     82 {
     83     Big_int ret;
     84     if(x < y) std::swap(x, y);
     85     ret.idx = x.idx;
     86     for(int i = 0; i < ret.idx; i++)
     87     {
     88         if(x.s[i] < y.s[i])
     89         {
     90             x.s[i] += 10;
     91             x.s[i + 1]--;
     92         }
     93         ret.s[i] = x.s[i] - y.s[i];
     94     }
     95     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
     96     return ret;
     97 }
     98 
     99 inline Big_int operator * (const Big_int x, const Big_int y)
    100 {
    101     Big_int ret;
    102     ret.idx = x.idx + y.idx;
    103     for(int i = 0; i < x.idx; i++)
    104         for(int j = 0; j < y.idx; j++)
    105         {
    106             ret.s[i + j] += x.s[i] * y.s[j];
    107             ret.s[i + j + 1] += ret.s[i + j] / 10;
    108             ret.s[i + j] %= 10;
    109         }
    110     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
    111     return ret;
    112 }
    113 
    114 inline Big_int operator / (Big_int x, int y)
    115 {
    116     int i, j, res = 0;
    117     Big_int ret;
    118     for(i = x.idx - 1; i >= 0; i--)
    119     {
    120         ret.s[i] = (x.s[i] + res * 10) / y;
    121         res = (x.s[i] + res * 10) % y;
    122     }
    123     ret.idx = x.idx;
    124     while(!ret.s[ret.idx - 1] && ret.idx > 1) ret.idx--;
    125     return ret;
    126 }
    127 
    128 inline Big_int Big(int x)
    129 {
    130     Big_int ret;
    131     ret = x;
    132     return ret;
    133 }
    134 
    135 inline Big_int max(const Big_int x, const Big_int y)
    136 {
    137     return x < y ? y : x;
    138 }
    139 
    140 inline bool cmp(node x, node y)
    141 {
    142     return x.a * x.b < y.a * y.b;
    143 }
    144 
    145 Big_int a, ans;
    146 
    147 int main()
    148 {
    149     int i;
    150     scanf("%d", &n);
    151     a = read();
    152     scanf("%d", &b);
    153     for(i = 1; i <= n; i++) scanf("%d %d", &p[i].a, &p[i].b);
    154     std::sort(p + 1, p + n + 1, cmp);
    155     for(i = 1; i <= n; i++)
    156     {
    157         ans = max(ans, a / p[i].b);
    158         a = a * Big(p[i].a);
    159     }
    160     ans.print();
    161     return 0;
    162 }
    View Code
  • 相关阅读:
    LeetCode: LRU Cache
    LeetCode: Reorder List
    LeetCode: Linked List Cycle I && II
    LeetCode: Word Break I && II
    LeetCode: Single Number I && II
    太坑了,mybatis注解一对多,id没了
    ajax请求参数的格式
    查询结果拼接
    id拼接保存到单个字段后作为表连接的查询条件
    seam的定时轮巡
  • 原文地址:https://www.cnblogs.com/zhenghaotian/p/6901854.html
Copyright © 2020-2023  润新知