C++博客-I believe-随笔分类-规律

joj 2243 Endless Carry

luis — Tue, 14 Jul 2009 09:18:00 GMT

作为2进制的加法，从k加1变成k+1可能会出现若干进位。如1011加1就会有2个进位。给定n，从0开始不停地加1直到n，计算在此过程中总共会有多少次进位。

Input

输入的每一行有单独的一个值n。n=0标志输入结束

Output

对于输入的每一行，使用单独一行输出对应结果。

Sample Input

Sample Output

1
3
8

总结规律：
1，0
2，1
3，1
4，3
5，3
6，4
7，4
8，7
9，7
10，8
11，8
12，10
16，15
32，31
64，63
你会发现2^n的都是2^n-1
其他13=8+4+1=(8-1)+(4-1)+(1-1),有多少个2^n,则减多少次1。
直接根据位运算即可求

luis 2009-07-14 17:18 发表评论

joj 2278 count the square

luis — Fri, 03 Jul 2009 08:41:00 GMT

There is chessboard which has N * M squares size of 1 * 1 and small squares can build up to a large one. Can you tell Mr. Guo that how many squares in the chessboard?

For Example, when N = 2 and M = 3, there are 6 squares size of 1 * 1 and 2 squares size of 2 * 2 in the chessboard, so the answer is 8.

Input

There are several lines in the input, each line consist of two positive integers N and M (1 <= N <= 100, 1 <= M <= 100) except the last line which N = 0 and M = 0 implying the end of the input and you should not process it.

Output

Please output the number of the squares for each chessboard in a single line.

Sample Input

Sample Output

1
8
14

启发：人是怎么算的，就让程序怎么算，一步一步推。
#include
using namespace std;
int main()
{
int n,m;
long sum=0;
int i;
while(cin>>n>>m,n)
{
if(n==m)
{
sum=n*(n+1)*(2*n+1)/6;
cout<}
else
{
for(i=0;isum+=(n-i)*(m-i);
cout<}
sum=0;
}
return 0;
}

luis 2009-07-03 16:41 发表评论

Round Up by Six巧妙转化

luis — Thu, 02 Jul 2009 16:56:00 GMT


In arithmetic we often use the scheme called 'round up by five' to round up a decimal fraction number into a decimal integer number whose fraction part equals to or is bigger than 0.5, otherwise, its fraction part is discarded.
In this problem, you are to implement a similar sheme called 'round up by six'.
Input
The first line of the input is an integer N indicating the number of cases you are to process. Then from the second line there are N lines, each containing a number (may or may not be fractional).
Output
For each number in the input, you are to print the round-up-by-six version of the number in a single line.
Sample Input
2
25.599
0.6

Sample Output
25
1

#include
#include
using namespace std;
int main()
{int n;
cin>>n;
while(n--)
{  double a;
cin>>a;
double b=floor(a);
if(a-b>=0.6)
cout<


luis 2009-07-03 00:56 发表评论

kmp算法实现及 kmp改进算法的实现笔记

luis — Tue, 30 Jun 2009 06:12:00 GMT

本文是结合刘大有主编《数据结构》所写笔记，如果想学习kmp.推荐http://blog.csdn.net/china8848/archive/2008/04/29/2342243.aspx
下面的估计我都看不懂
#include
#include
#include
using namespace std;
int f[100];
void createf(string pat)//find数组
{
memset(f,0,sizeof(f));
int m=pat.size();
f[0]=-1;
for(int j=1;j {
  int i=f[j-1];
  while(pat[j]!=pat[i+1]&&i>=0)//只要p[j]与p[i+1]不等，则递推计算i
  i=f[i];
  if(pat[j]==pat[i+1])
  f[j]=i+1;
  else
  f[j]=-1;
}
}
/*                        已知f[13]=6;acbacbcacbacb,求f[14]acbacbcacbacba则先判断pat[6+1]是否等于pat[14]，不等
                                再判断 i=f[6]acbacb=3;判断pat[14]pat[3+1]相等，则i=3
                           可得f[j]=3+1;

*/
int kmp(string pat,string str)//求pat在str中的第一个匹配的起点
{
int p=0,s=0;
createf(pat);
int m=pat.size();
int n=str.size();
while(p {
  if(pat[p]==str[s])
  {
   p++;s++;
  }
  else
  {
  if(p==0)s++;
  else
  p=f[p-1]+1;//s无需变化,举例，abababababba
/*                                                          |
                                                             s指向a
                                                     ababb
//                                                           |
//                                                           p指向b,s和p失配

                                                    abababababba
//                                                          |
                                                            s s不变
  由于ab=ab，既f[p-1]=2               ababb
//     p=f[p-1]+1=3                                |
//                                                          p 回到f[p-1]+1的位置;

相等，则s和p都加1.
//
*/
}
}
     if(p     return -1;
     return s-m;
}

int main()
{
//freopen("s.txt","r",stdin);
//freopen("key.txt","w",stdout);
cout< getchar();
//system("PAUSE");
return   0;
}

改进的kmp
----------------------以下是原kmp生成next值的算法---------------------
void createf(string pat)//find数组
{
memset(f,0,sizeof(f));
int m=pat.size();
f[0]=-1;
for(int j=1;j {
  int i=f[j-1];
  while(pat[j]!=pat[i+1]&&i>=0)//只要p[j]与p[i+1]不等，则递推计算i
  i=f[i];
  if(pat[j]==pat[i+1])
  f[j]=i+1;
  else
  f[j]=-1;
}
}
/*                        已知f[13]=6;acbacbcacbacb,求f[14]acbacbcacbacba则先判断pat[6+1]是否等于pat[14]，不等
                                再判断 i=f[6]acbacb=3;判断pat[14]pat[3+1]相等，则i=3
                           可得f[j]=3+1;

*/
-------------------以下还是未改进kmp------------
void get_next(String T,int &next[])
{
i=1;j=0;next[1]=0;
while(i<=T.Length)
{
    if(j==0 || T[i]==T[j]){++i;++j; next[i]=j;/**********(2)*/}
    else j=next[j];
}
}

--------------------以下是改进型----------
上面的f[j]相当于next[]
int i,j,len,n;
        len=strlen(s);
        //KMP构造
        i=0,j=-1;
        next[0]=-1;
        while(i       {
            if(j==-1||s[i]==s[j])
           {
                i++,j++;
                if(s[i]!=s[j])
                   next[i]=j;
                else
                   next[i]=next[j];
            //    printf("next[%d]=%d\n",i,next[i]);
            }
           else j=next[j];
        }

luis 2009-06-30 14:12 发表评论

joj 2344 Dissolution of Integer

luis — Tue, 30 Jun 2009 02:49:00 GMT

As you known,a positive integer can be written to a form of products of several positive integers(N = x1 * x2 * x3 *.....* xm (xi!=1 (1<=i<=m))).Now, given a positive integer, please tell me how many forms of products.

Input

Input consists of several test cases.Given a positive integer N (2<=N<=200000) in every case.

Output

For each case, you should output how many forms of the products.

Sample Input

12
100
5968

Sample Output

4
9
12

#include
#include
using namespace std;

int main()
{
int a[200001];
freopen("s.txt","r",stdin);
freopen("key.txt","w",stdout);
int i,j,n;
memset(a,0,sizeof(a));
a[1]=1;
for(i=1;i<=200000;i++)
  for(j=1;i*j<=200000;j++)
   a[i*j]+=a[j];
while(scanf("%d",&n)!=EOF)
  printf("%d\n",a[n]/2);
//system("PAUSE");
return   0;
}

启发：申请数组一定要大一号a[200001]，否则无法清空a[200000];
发现递推规律要从特点和出发，乘法自然是乘法。

luis 2009-06-30 10:49 发表评论

C++博客-I believe-随笔分类-规律

joj 2243 Endless Carry

Input

Output

Sample Input

Sample Output

joj 2278 count the square

Input

Output

Sample Input

Sample Output

Round Up by Six巧妙转化

Input

Output

Sample Input

Sample Output

kmp算法实现 及 kmp改进算法的实现笔记

joj 2344 Dissolution of Integer

Input

Output

Sample Input

Sample Output

kmp算法实现及 kmp改进算法的实现笔记