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  • HDOJ(~1004)

    T1000

    1 #include <stdio.h>
    2 int main()
    3 {
    4     int a, b;
    5     while (scanf("%d %d", &a, &b) != EOF)
    6         printf("%d
    ", a + b);
    7     return 0;
    8 }

    T1001

     1 #include <stdio.h>
     2 #ifdef AUTHOR
     3 Code By Hatsuakira Tenan
     4 #endif
     5 int f(int n)
     6 {
     7     int js = 0;
     8     for (; n > 0; --n)
     9         js += n;
    10     return js;
    11 }
    12 int main()
    13 {
    14     int n;
    15     while (scanf("%d", &n) != EOF)
    16         printf("%d
    
    ", f(n));
    17     return 0;
    18 }

    T1002

     1 #include <stdio.h>
     2 #include <string.h>
     3 #include <stdlib.h>
     4 #include <algorithm>
     5 #define MAXLENGTH 1002
     6 #define MAXCARRY 10000
     7 #ifdef AUTHOR
     8 Code By Hatsuakira Tenan
     9 #endif
    10 struct largeint
    11 {
    12     int num[MAXLENGTH], len;
    13     largeint()
    14     {
    15         len = 1;
    16         memset(num, 0, sizeof(num));
    17     }
    18 };
    19 void scanf_override(largeint &input)
    20 {
    21     int tmplen;
    22     char tmp[MAXLENGTH];
    23     scanf("%s", tmp);
    24     std::reverse(tmp, tmp + (tmplen = strlen(tmp)));
    25     input.len = tmplen / 4 + (tmplen % 4 > 0);
    26     for (int i = 0, j = 0; i < tmplen; ++i, ++j)
    27     {
    28         input.num[j] = tmp[i] - '0';
    29         if (tmp[++i] == '')
    30             break;
    31         input.num[j] += 10 * (tmp[i] - '0');
    32         if (tmp[++i] == '')
    33             break;
    34         input.num[j] += 100 * (tmp[i] - '0');
    35         if (tmp[++i] == '')
    36             break;
    37         input.num[j] += 1000 * (tmp[i] - '0');
    38     }
    39 }
    40 inline largeint operator +(largeint a, largeint b)
    41 {
    42     int tmp = 0;
    43     a.len = std::max(a.len, b.len);
    44     for (int i = 0; i < a.len; ++i)
    45     {
    46         a.num[i] += b.num[i] + tmp;
    47         tmp = a.num[i] / MAXCARRY;
    48         a.num[i] %= MAXCARRY;
    49     }
    50     if (tmp > 0)
    51         a.num[a.len++] = 1;
    52     return a;
    53 }
    54 void printf_override(largeint output)
    55 {
    56     printf("%d", output.num[--output.len]);
    57     for (int i = output.len - 1; i > -1; --i)
    58         printf("%04d", output.num[i]);
    59 }
    60 void work(int k)
    61 {
    62     largeint a = largeint(), b = largeint();
    63     scanf_override(a);
    64     scanf_override(b);
    65     printf("Case %d:
    ", k + 1);
    66     printf_override(a);
    67     printf(" + ");
    68     printf_override(b);
    69     printf(" = ");
    70     a = a + b;
    71     printf_override(a);
    72     printf("
    ");
    73 }
    74 int main()
    75 {
    76     int T;
    77     scanf("%d", &T);
    78     work(0);
    79     for (int i = 1; i < T; ++i)
    80         { printf("
    "); work(i); }
    81     return 0;
    82 }

     T1003

    最长连续子序列和, 根据数据范围(N<=100, 000 & T<=20)选择使用O(n)的算法, 是较为基础的动态规划.

     1 #include <stdio.h>
     2 #include <limits.h>
     3 #define MAXLENGTH 100005
     4 #ifdef AUTHOR
     5 Code By Hatsuakira Tenan
     6 #endif
     7 int number[MAXLENGTH], start, end, max;
     8 void dp(int length)
     9 {
    10     int max_tmp = 0, start_tmp = 0, end_tmp = 0;
    11     max_tmp = max = INT_MIN;
    12     for (int i = 0; i < length; ++i)
    13     {
    14         if (max_tmp < 0)
    15         {
    16             if (number[i] > max_tmp)
    17             {
    18                 max_tmp = number[i];
    19                 start_tmp = end_tmp = i;
    20                 if (max < max_tmp)
    21                 {
    22                     max = max_tmp;
    23                     start = start_tmp;
    24                     end = end_tmp;
    25                 }
    26             }
    27         }
    28         else
    29         {
    30             max_tmp += number[i];
    31             end_tmp = i;
    32             if(max_tmp > max)
    33             {
    34                 max = max_tmp;
    35                 start = start_tmp;
    36                 end = end_tmp;
    37             }
    38         }
    39     }
    40 }
    41 void work(int R)
    42 {
    43     printf("Case %d:
    ", R);
    44     int n;
    45     scanf("%d", &n);
    46     for (int i = 0; i < n; ++i)
    47         scanf("%d", &number[i]);
    48     dp(n);
    49     printf("%d %d %d
    ", max, start + 1, end + 1);
    50 }
    51 int main()
    52 {
    53     int T;
    54     scanf("%d", &T);
    55     for (int i = 1; i < T; printf("
    "), ++i)
    56         work(i);
    57     work(T);
    58     return 0;
    59 }

     T1004 用STL 做这个题很耍赖, 但是鉴于数据量不大所以直接暴力遍历也是可以的.

     1 #include <stdio.h>
     2 #include <string.h>
     3 #include <map>
     4 #include <string>
     5 using std::map;
     6 using std::string;
     7 map<string, int> set;
     8 void work(int n)
     9 {
    10     set.clear();
    11     int maxn = -1;
    12     string tmp_stl_string;
    13     char tmp_string[21];
    14     for (int i = 0; i < n; tmp_stl_string = tmp_string, ++set[tmp_stl_string], ++i)
    15         scanf("%s", tmp_string);
    16     for (map<string, int>::iterator it = set.begin(); it != set.end(); ++it)
    17         if (maxn < it -> second)
    18         {
    19             maxn = it -> second;
    20             tmp_stl_string = it -> first;
    21         }
    22     printf("%s
    ", tmp_stl_string.c_str());
    23 }
    24 int main()
    25 {
    26     int n;
    27     scanf("%d", &n);
    28     while (n > 0)
    29     {
    30         work(n);
    31         scanf("%d", &n);
    32     }
    33     return 0;
    34 }
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  • 原文地址:https://www.cnblogs.com/hatsuakiratenan/p/3362962.html
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