ACM-data computing, acm computing

ACM-data computing, acm computing
Number calculation-(recursion (timeout) and non-recursion) time limit (Common/Java): 1000 MS/3000 MS running memory limit: 65536 KByte
Total submissions: 1050 pass the test: 312

Description

It is required to find the number of numbers with the following properties (including the natural number n of the input ):
Enter a natural number n (n <= 1000), and then process the natural number as follows:
1. No processing;
2. Add a natural number to its left, but the natural number cannot exceed half of the original number;
3. After the number is added, continue to process according to this rule until the natural number cannot be added.

Input

A natural number n

Output

One number, indicating the number of conditions

Sample Input

6

Sample output

6

Prompt

Example: The number of conditions is 6, 16, 26,126, 36,136

Question Source

NOIP2001 popularity Group

1 # include <iostream> 2 using namespace std; 3 static int sum = 1; 4 static int arr [1000] = {0}; 5 void fun (int & k) // Recursive Method 6 {7 if (k = 0) 8 return; 9 for (int I = 1; I <= k; I ++) 10 {11 sum ++; 12 int k2 = I/2; 13 fun (k2); 14} 15} 16 17 int f (int & k) {// use a non-recursive method to save the computation result 18 int count = 0; 19 if (k = 0) 20 return 0; 21 int I; 22 for (I = 1; I <= k; I ++) {23 if (arr [I]! = 0) {24 count + = arr [I]; 25} 26 else {27 int k2 = I/2; 28 arr [I] = f (k2) + 1; 29} 30} 31 if (count! = 0) 32 return count; 33 return arr [k]; 34} 35 int main () 36 {37 int n1; 38 cin> n1; 39 int k = n1/2; 40 fun (k); 41 cout <sum <endl; 42 cout <f (n1) <endl; 43 return 0; 44}

Http://acm.njupt.edu.cn/acmhome/problemdetail.do? & Method = showdetail & id = 1010

 
Acm enters a series of data computing and 12345 until the end of the file (what does this mean)

The input end with a file. This is a common input method in ACM.

# Include <stdio. h>
Int main ()
{
Double;
Double sum;
Sums = 0.0;
While (scanf ("% lf", & )! = EOF)
Sum + =;
Printf ("%. 4f \ n", sum );
Return 0;
}

Calculation of numbers in strings using C language ACM simple question

# Include <stdio. h>
# Include <string. h>
Int main ()
{
Int cases, len;
Int I;
Int ans;
Char str [100];
Scanf ("% d", & cases );
While (cases --)
{
Ans = 0;
Scanf ("% s", str );
Len = strlen (str );
For (I = 0; I <len; ++ I)
{
If (str [I]> = '0' & str [I] <= '9 ')
Ans ++;
}
Printf ("% d \ n", ans );
}
Return 0;
}

This is my program code, and the AC has passed.
Your program has some bad habits.
Hope this will be useful to you