codeforces round 26前两题水题

水题，没必要往这儿放的，只是保存一下而已 ，哎，后边的就不会写了……郁闷

A. Almost Prime

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and n, inclusive.

Input

Input contains one integer number n (1 ≤ n ≤ 3000).

Output

Output the amount of almost prime numbers between 1 and n, inclusive.

Sample test(s)

Input

10

Output

2

Input

21

Output

8
#include <stdio.h>#include <string.h>#include <math.h>int p[2000];int ap(int n){int count = 1,i;int a[50] = {0};for (i=0;n>=p[i];i++){if (n%p[i]==0){n /= p[i];a[count] = p[i];if (a[count] != a[count-1])count++;i--;}}if (count==3)return 1;return 0;}int main(void){int n,count,i,j,a[3600];memset (a,1,sizeof(a));for (i=2;i<sqrt(3500);i++)if(a[i])for (j=2;i*j<3500;j++)a[i*j] = 0;for (i=2,j=0;i<3500;i++)if (a[i]){p[j] = i;j++;}count = 0;scanf("%d",&n);for (i=6;i<=n;i++)if(ap(i))count++;printf("%d/n",count);return 0;} 

B. Regular Bracket Sequence

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, sequences "(())()", "()" and "(()(()))" are regular, while ")(", "(()" and "(()))(" are not.

One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?

Input

Input consists of a single line with non-empty string of "(" and ")" characters. Its length does not exceed 10⁶.

Output

Output the maximum possible length of a regular bracket sequence.

Sample test(s)

Input

(()))(

Output

4

Input

((()())

Output

6

#include <stdio.h>char a[1000010];int main(void){int i=1,t=0,count=0;char c;while (c = getchar() && c != '/n'){if (c=='('){i++;a[i] = c;}else if (c==')'&&a[i] =='C'){count++;i--;}}printf("%d/n",count*2);return 0;}