hdu4279 Number

Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2284 Accepted Submission(s): 632

Problem Description

　　Here are two numbers A and B (0 < A <= B). If B cannot be divisible by A, and A and B are not co-prime numbers, we define A as a special number of B.
　　For each x, f(x) equals to the amount of x’s special numbers.
　　For example, f(6)=1, because 6 only have one special number which is 4. And f(12)=3, its special numbers are 8,9,10.
　　When f(x) is odd, we consider x as a real number.
　　Now given 2 integers x and y, your job is to calculate how many real numbers are between them.

Input

　　In the first line there is an integer T (T <= 2000), indicates the number of test cases. Then T line follows, each line contains two integers x and y (1 <= x <= y <= 2^63-1) separated by a single space.

Output

　　Output the total number of real numbers.

Sample Input

21 11 10

Sample Output

04
Hint
 For the second case, the real numbers are 6,8,9,10. 
 

Source

2012 ACM/ICPC Asia Regional Tianjin Online

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暴力打表找规律

#include <iostream>#include <stdio.h>#include <string.h>#include <math.h>using namespace std;__int64 num[15]={0,0,0,0,0,0,1,1,2,3,4,4,5};__int64 get(__int64 x){    if(x<=12)return num[x];    __int64 temp=(__int64)sqrt((double)x);    if(temp&1)        return x/2-1;    else        return x/2-2;    return -1;}int main(){    int tcase;    __int64 m,n;    scanf("%d",&tcase);    while(tcase--)    {        scanf("%I64d%I64d",&n,&m);        n--;        __int64 p1=get(n);        __int64 p2=get(m);        printf("%I64d\n",p2-p1);    }    return 0;}