C++博客-雪黛依梦-随笔分类-简单题

HDU 1061

雪黛依梦 — Sat, 11 Sep 2010 14:00:00 GMT

//解题思路：求一个数最右边的一位数，即%10，
//故：N^N 的最右边的一位数是( N^N ) % 10,根据同余定理= (N % 10)^N
//通过观察由 1 - 9的 N 次幂容易找出相应的规律,
#include
#include
using namespace std;
#include

int main ()
{
    int t;
    scanf ("%d", &t);
    while ( t -- )
    {
          int n, a;
          scanf ("%d", &n);
          a = n % 10;

          if ( a == 0 )
          printf ("%d\n", 0);

          else if ( a == 1 )
          printf ("%d\n", 1);

          else if ( a == 5 )
          printf ("%d\n", 5);

          else if ( a == 6 )
          printf ("%d\n", 6);

          else if ( a == 2 )
          {
               if ( n % 4 == 0 )
               printf ("%d\n", 6);

               else if ( n % 4 == 1 )
               printf ("%d\n", 2);

               else if ( n % 4 == 2 )
               printf ("%d\n", 4);

               else if ( n % 4 == 3 )
               printf ("%d\n", 8);
          }

          else if ( a == 3 )
          {
               if ( n % 4 == 0 )
               printf ("%d\n", 1);

               else if ( n % 4 == 1 )
               printf ("%d\n", 3);

               else if ( n % 4 == 2 )
               printf ("%d\n", 9);

               else if ( n % 4 == 3 )
               printf ("%d\n", 7);
          }

          else if ( a == 4 )
          {
               if ( n % 2 == 0 )
               printf ("%d\n", 6);

               else if ( n % 2 == 1 )
               printf ("%d\n", 4);
          }

          else if ( a == 7 )
          {
               if ( n % 4 == 0 )
               printf ("%d\n", 1);

               else if ( n % 4 == 1 )
               printf ("%d\n", 7);

               else if ( n % 4 == 2 )
               printf ("%d\n", 9);

               else if ( n % 4 == 3 )
               printf ("%d\n", 3);
          }

          else if ( a == 8 )
          {
               if ( n % 4 == 0 )
               printf ("%d\n", 6);

               else if ( n % 4 == 1 )
               printf ("%d\n", 8);

               else if ( n % 4 == 2 )
               printf ("%d\n", 4);

               else if ( n % 4 == 3 )
               printf ("%d\n", 2);
          }

          else if ( a == 9 )
          {
               if ( n % 2 == 0 )
               printf ("%d\n", 1);

               else if ( n % 2 == 1 )
               printf ("%d\n", 9);
          }
    }
   system ("pause");
   return 0;
}

雪黛依梦 2010-09-11 22:00 发表评论

HDU 1058

雪黛依梦 — Sat, 11 Sep 2010 12:56:00 GMT

//重在理解方法：每次找到最小的 2 3 5  7 的因子数，之后利用所存数的下标的关系改变
#include
using namespace std;

int num[5843];  //存储前5842个丑数

int find_min ( int a, int b, int c, int d )
{
    int temp = a < b ? a : b;
    int index = c < d ? c : d;
    return temp < index ? temp : index;
}

void solve ( )
{
     int i1, i2, i3, i4, i;
     int h1, h2, h3, h4;
     i1 = i2 = i3 = i4 = 1;
     num[1] = 1;
     for (i = 2; i < 5843; i ++ )
     {
         h1 = num[i1] * 2;
         h2 = num[i2] * 3;
         h3 = num[i3] * 5;
         h4 = num[i4] * 7;
         int min = find_min ( h1, h2, h3, h4 );

         num[i] = min;


         //易错点：这里不可以用else if 因为ti中可能会有相同的最小值，如当：min = 6 时
         if ( min == h1 )
         i1 ++;

         if ( min == h2 )
         i2 ++;

         if ( min == h3 )
         i3 ++;

         if ( min == h4 )
         i4 ++;
     }
}

int main ()
{
    solve ();
    int n;
    while ( scanf ("%d", &n), n )
    {
          if ( n % 100 != 11 && n % 10 == 1 )
            printf ("The %dst humble number is %d.\n", n, num[n]);

          else if ( n % 100 != 12 && n % 10 == 2 )
            printf ("The %dnd humble number is %d.\n", n, num[n]);

          else if ( n % 100 != 13 && n % 10 == 3 )
            printf ("The %drd humble number is %d.\n", n, num[n]);

          else
            printf ("The %dth humble number is %d.\n", n, num[n]);
    }
    //system ("pause");
    return 0;
}

雪黛依梦 2010-09-11 20:56 发表评论

HDU 1056

雪黛依梦 — Tue, 07 Sep 2010 13:04:00 GMT

一个水题可是错了好几次，原因就是在scanf n 时输入的格式错了，应该是：%lf

//思路：首先是通过打表找到当长度取 0.01到5.20之间的数时，最多要多少张card，
//然后将张数存到length中用下表i表示张数
#include
#include
using namespace std;
#include

int main ()
{
    double length[300];
    memset ( length, 0, sizeof(length));
    for ( int i = 1; i <= 300; i ++)
    {
        for (int j = 1; j <= i; j ++)
        {
            length[i] += 1.0 / ( j + 1 );
        }
        //printf ("%d %.2f\n", i, length[i]);
    }

    double n;
    while ( scanf ("%lf", &n) && n != 0.00 )
    {
          for ( int i = 1;i <= 300; i ++ )
          {
              if (  n <= length[i] )
              {
                    printf ("%d card(s)\n", i);
                    break;
              }
          }
    }
    system ("pause");
    return 0;
}

雪黛依梦 2010-09-07 21:04 发表评论

hdu 1048

雪黛依梦 — Tue, 07 Sep 2010 03:02:00 GMT

一水题也被我WA 了好几次哦！就是开头的那个必须是START
所以必须多加一个strcmp ( first, ''START'')

#include
#include
using namespace std;

int main ()
{
    char first[11];
    char last[8];
    char str[201];

    while (  scanf ("%s", &first) && strcmp( first, "EDNOFINPUT") && !strcmp (first, "START") )
    {
         getchar ();
         int len = 0;
         while ( (str[len] = getchar ()) != '\n')
         len ++;

         for ( int i = 0; i < len; i ++ )
         {
             if ( str[i] >= 'A' && str[i] <= 'E')
             {
                  printf ("%c", str[i] + 21);
             }
             else if ( str[i] >= 'F' && str[i] <= 'Z')
             {
                  printf ("%c", str[i] - 5);
             }
             else
             printf ("%c", str[i]);
         }
         printf ("\n");
         scanf ("%s", last);
    }
   //system ("pause");
   return 0;
}

雪黛依梦 2010-09-07 11:02 发表评论

hdu 1040

雪黛依梦 — Mon, 06 Sep 2010 13:19:00 GMT

#include
#include
using namespace std;

int main ()
{
    int a[1005];
    int t;
    scanf ( "%d", &t );
    while ( t -- )
    {
          int n;
          scanf ( "%d", &n );
          for ( int i = 0; i < n; i ++ )
          {
              scanf ("%d", &a[i]);
          }
          sort ( a, a + n );

          printf ("%d", a[0]);
          for ( int i = 1; i < n; i ++ )
          {
              printf (" %d",a[i]);
          }
          printf ("\n");
    }
   //system ("pause");
   return 0;
}

简单的排序题，直接利用sort默认的排序规则：升序排列

雪黛依梦 2010-09-06 21:19 发表评论

hdu 1021

雪黛依梦 — Mon, 06 Sep 2010 13:06:00 GMT

为啥少了一个else 就错了呢？

//显然数据范围非常的大，测试a[10000]时已经无法正确的表示了
//要用用相关的知识进行处理 : ( a + b ) % 3 = ( a % 3 + b % 3) % 3
#include
#include
#include

int a[1000009];

int main ()
{
    int n;
    a[0] = 7 % 3;
    a[1] = 11 % 3;
    for ( int i = 2; i < 1000009; i ++ )
    {
        a[i] = ( a[i - 1] + a[i - 2] ) % 3;
    }

    while ( scanf ("%d", &n) != EOF )
    {
          if ( !a[n] )
             printf ("yes\n");
          else
           printf ("no\n");
    }
    //system ("pause");
    return 0;
}

雪黛依梦 2010-09-06 21:06 发表评论

HDU 1017

雪黛依梦 — Sat, 04 Sep 2010 05:26:00 GMT

#include
#include
#include
#include
int main ()
{
          int N;
          scanf ("%d", &N );
          for ( int i = 0; i < N; i ++ )
          {
              //输出之间存在空行
              if (i)
              printf ("\n");

              int m, n;
              int num = 0;    //记录看这是第几组数
              while ( scanf ("%d %d", &n, &m), m || n )           //水题也WA 了好多次就是这里错了
              {
                    num ++;
                    int count = 0;    //用于记录满足条件的 a  b 整数对
                    for ( int a = 1; a < n; a ++ )    //  b
                    {
                        for ( int b = a + 1; b < n; b ++ )  // a
                        {
                            if ( (a * a + b * b + m) % ( a * b ) == 0 )
                            count ++;
                        }
                    }
                    printf ("Case %d: %d\n", num, count);
              }

          }

   //system ("pause");
   return 0;
}

雪黛依梦 2010-09-04 13:26 发表评论

HDU 1014

雪黛依梦 — Fri, 03 Sep 2010 12:50:00 GMT

水题：只是格式好容易错的，像那个是从25 行输出，我的就变成了26

#include
#include
#include
#include

int a[100000];
int main ()
{
    int step, mod;

    while ( scanf ("%d %d", &step, &mod) != EOF )
    {
          a[0] = 0;
          a[1] = ( a[0] + step) % mod;
          int k = 1;
          while ( a[k] != 0 )
          {
                k ++;
                a[k] = ( a[k - 1] + step ) % mod;
          }

          if ( k == mod )
          printf ("%10d%10d    Good Choice\n", step, mod);
          else
          printf ("%10d%10d    Bad Choice\n",step, mod);
          printf ("\n");
    }
    //system ("pause");
    return 0;
}

雪黛依梦 2010-09-03 20:50 发表评论

HDU 1012

雪黛依梦 — Fri, 03 Sep 2010 11:15:00 GMT

#include
#include
#include
#include

int f ( int n )
{
    int product = 1;
    for ( int i = 1; i <= n; i ++ )
    {
        product *= i;
    }
    return product;
}

int main ()
{
    double a[10];
    a[0] = 1;
    for ( int i = 1; i < 10; i ++ )
    {
        a[i] = 1.0 / f (i);  //求阶层的倒数
    }

    double sum[10];
    sum[0] = 1;
    for ( int i = 1; i < 10; i ++ )
    {
        sum[i] = sum[i - 1] + a[i];   //求和
    }


    printf ("n e\n");
    printf ("- -----------\n");
    printf ("0 1\n");
    printf ("1 2\n");
    printf ("2 2.5\n");
    for ( int i = 3; i < 10; i ++)
    {
        printf ("%d %.9f\n", i, sum[i]);
    }
    system ("pause");

   return 0;
}

雪黛依梦 2010-09-03 19:15 发表评论

HDU 1008

雪黛依梦 — Fri, 03 Sep 2010 10:37:00 GMT

#include
#include
#include
#include
int main ()
{
    int a[1000];
    int n;
    while ( scanf ("%d", &n) && n )
    {
          a[0] = 0;
          for ( int i = 1; i <= n; i ++ )
          {
              scanf ("%d", &a[i]);
          }

          int sum = 0;
          for ( int i = 1; i <= n; i ++ )
          {
              if ( a[i] > a[i - 1] )
              {
                   sum += (a[i] - a[i - 1]) * 6 + 5;
              }
              else if ( a[i] < a[i - 1])
              {
                   sum += (a[i - 1] - a[i]) * 4 + 5;
              }
              else
              {
                  sum += 5;
              }
          }

          printf ("%d\n", sum);
    }
   system ("pause");
   return 0;
}

雪黛依梦 2010-09-03 18:37 发表评论

HDU 1279

雪黛依梦 — Sat, 28 Aug 2010 03:27:00 GMT

用hash 怎么做呢？

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

int main ()
{
    int t;
    int n;
    while ( scanf ("%d", &t) != EOF )
    {
          for ( int i = 0; i < t; i ++ )
          {
              scanf ( "%d", &n );
              bool flag = 0;
              int count = 0;
              while ( n != 1 )
              {
                    if ( n % 2 == 0 )
                    {
                         n = n / 2;

                    }
                    else if ( n % 2 != 0 )
                    {
                         count ++;
                         flag = 1;
                         count == 1 ? printf ("%d", n) : printf (" %d", n);
                         n = n * 3 + 1;

                    }
              }
              if ( !flag )
              printf ("No number can be output !\n");
              else
              printf ("\n");
          }
    }

    // system ("pause");
     return 0;
}

雪黛依梦 2010-08-28 11:27 发表评论

hdu 1004

雪黛依梦 — Tue, 24 Aug 2010 08:46:00 GMT

开始时这道题目想了挺久的，也走了一点弯路，其实主要是C 语言的二维字符数组没理解好，都是一些基础知识没有掌握，注意打红色的部分就好了

1#include <stdio.h>
2#include <stdlib.h>
3#include <string.h>
4char code[1001][17];
5int num[1001];
6int main ()
7{
8  int n;
9  while (scanf ("%d", &n) != EOF && n != 0)
10  {
11   getchar ();
12   memset (code, 0, sizeof (code));
13   memset (num, 0 , sizeof (num));
14
15   for (int i = 0; i < n; i ++)    //输入处理
16   {
17    int j = 0;
18    while ( (code[i][j] = getchar ()) !=  '\n' )     i 代表这是第几个串
19    j ++;
20   }
21
22
23   for (int m = 0; m < n; m  ++)     //匹配
24   {
25    for (int k = m + 1; k < n; k ++)
26    {
27     if ( !strcmp (code[m], code[k]) )  //相等返回 0
28      num[m]++;
29    }
30   }
31
32   int max = -1;   int index = 0;
33   for ( int k = 0; k < n; k ++)  //找到出现次数最多的下标
34   {
35       if (max < num[k])
36       {
37         max = num[k];
38         index = k;
39       }
40   }
41   printf ("%s", code[index]);
42 }
43 //system ("pause");
44 return 0;
45}
46

雪黛依梦 2010-08-24 16:46 发表评论

hdu 1005

雪黛依梦 — Tue, 24 Aug 2010 06:01:00 GMT

//本来以为是一道简单题谁知道。。。。
//用递归写想都不要想，绝对超时，所以还应当回到题目本身的分析上来
//正确思路是：因为mod7的关系，而且f(1)=f(2)=1，所以f(n)的值是循环分布的，而且一定会回到f(n-1)=f(n)=1，
//并且还可得出，这个循环不大于49，因为相邻连个f只有7种取值，这样f(n-1)和f(n)共有49种组合。
//所以，只要找出循环因子即可，寻找方法正是根据f(n-1)=f(n)再次出现的地方来计算
//可以首先为这个题目写一个测试的程序设定一个 a b n（n 比较小时）的值看看输出规律

1#include <stdio.h>
2#include <stdlib.h>
3int f[51];
4int main ()
5{
6    int a, b, n;
7    while ( scanf ("%d %d %d", &a , &b, &n) != EOF && a != 0 && b != 0 && n != 0 )
8    {
9          f[1] = f[2] = 1;
10          int i;
11          for (i = 3; i < 51; i ++)
12          {
13              f[i] = (a * f[i - 1] + b * f[i - 2]) % 7;
14              if ( f[i] == 1 &&  f[i - 1] == 1 )   //找到循环因子 i
15              {
16                   break;
17              }
18          }
19
20          n = n % (i - 2);
21          if (n == 0)   //刚好经过一个循环
22          printf ("%d\n", f[i - 2]);   //开始时，我是因为看了测试程序，把这里设定为输出 0 这种想法是错的，太片面了，因为数据范围很大
23          else
24          printf ("%d\n", f[n]);
25    }
26    //system ("pause");
27    return 0;
28}
29
30

雪黛依梦 2010-08-24 14:01 发表评论

hdu 1555

雪黛依梦 — Tue, 24 Aug 2010 04:31:00 GMT

1#include <stdio.h>
2int main ()
3{
4    int m, k;
5    while ( scanf ("%d %d", &m, &k) != EOF && m != 0 && k != 0)
6    {
7        int days = 0;
8        while ( m )
9        {
10            m --;
11            days ++;
12            if (days % k == 0)
13                m ++;
14        }
15
16        printf ("%d\n", days);
17    }
18    return 0;
19}
20

雪黛依梦 2010-08-24 12:31 发表评论

hdu 1032

雪黛依梦 — Mon, 23 Aug 2010 11:42:00 GMT

这样的水题也被我WA了几次，惭愧啊！就是就是没有注意当i > j 时输出位置不变的问题
看来做题目一定要细心，小心陷阱
我的代码：

1#include <stdio.h>
2#include <stdlib.h>
3
4int cylen (int a, int b)
5{
6    int max = 0;
7    for (int i = a; i <= b; i ++)
8    {
9        int n = i;
10        int count = 1;
11        while ( n != 1)
12        {
13              count ++;
14              if ( n % 2 == 0)
15              n = n / 2;
16              else
17              n = 3 * n + 1;
18        }
19
20        if (max < count)
21        max = count;
22    }
23        return max;
24}
25
26int main ()
27{
28    int i, j;
29    while ( scanf ("%d %d", &i, &j) != EOF )
30    {
31          int index;
32          if (i > j)
33          {
34                index = i;
35                i = j;
36                j = index;
37                printf ("%d %d %d\n", j, i, cylen (i, j));
38          }
39          else
40          printf ("%d %d %d\n", i, j, cylen (i, j));
41    }
42
43    //system ("pause");
44    return 0;
45}
46

网上的代码：用递归做的，两个代码复杂度完全一样

1# include <stdio.h>
2#include <stdlib.h>
3int fun(long  a ,int len)
4{
5    if (a ==1)
6        return len;
7    else if (a %2 ==0)
8        return fun(a/2,len+1);
9    else  return fun(3*a+1,len +1);
10}
11int main()
12{
13    unsigned long  a,b;
14    unsigned long i,max,temp;
15
16    while (scanf("%ld%ld",&a,&b)!=EOF)
17    {
18        printf("%ld %ld ",a,b);
19        if(a>b) {i = a;a=b;b=i;}
20        for (max = 0,i = a; i <= b; i ++)
21        {
22            temp = fun(i,1);
23            if (max < temp)
24                max = temp;
25        }
26        printf("%ld\n",max);
27    }
28    //system ("pause");
29    return 0;
30}
31

雪黛依梦 2010-08-23 19:42 发表评论