HNU2014暑假赛七

Quite Good NumbersTime Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KBTotal submit users: 76, Accepted users: 56Problem 12876 : No special judgementProblem description

A "perfect" number is an integer that is equal to the sum of its divisors (where 1 is considered a divisor). For example, 6 is perfect because its divisors are 1, 2, and 3, and 1 + 2 + 3 is 6. Similarly, 28 is perfect because it equals 1 + 2 + 4 + 7 + 14.
A "quite good" number is an integer whose "badness" ? the absolute value of the difference between the sum of its divisors and the number itself ? is not greater than a specified value. For example, if the allowable badness is set at 2, there are 11 "quite good" numbers less than 100: 2, 3, 4, 6, 8, 10, 16, 20, 28, 32, and 64. But if the allowable badness is set at 0 (corresponding to the "perfect" numbers) there are only 2: 6 and 28.
Your task is to write a program to count how many quite good numbers (of a specified maximum badness) fall in a specified range.

Input

Input will consist of specifications for a series of tests. Information for each test is a single line containing 3 integers with a single space between items:
• start (2 <= start < 1000000) specifies the first number to test
• stop (start <= stop < 1000000) specifies the last number to test
• badness (0 <= badness < 1000) specifies the maximum allowable badness
A line containing 3 zeros terminates the input.

Output

Output should consist of one line for each test comprising the test number (formatted as shown) followed by a single space and the number of values in the test range with badness not greater than the allowable value.

Sample Input

2 100 22 100 01000 9999 30 0 0

Sample Output

Test 1: 11Test 2: 2Test 3: 6

Problem Source

HNU Contest 

当时做这道题的时候，想的是一个一去求出每个数的差距badness，结果一个TLE，各种素数筛选优化都不行，最后看了别人的思路，才知道，可以这样去筛选。

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<fstream>
#include<algorithm>
#include<string>
#include<stack>
#include<queue>
#include<climits>
#include<map>
#define MAXP(x,y)  (((x)>(y))?(x):(y))
#define MINP(x,y)  (((x)<(y))?(x):(y))
const int MAX=0xfffffff;
using namespace std;
const int mx=1000010;
int a[1000010];
void init( )
{
    for(int i=2;i<mx;i++)
        a[i]=1;
    for(int i=2;i<=1000000;i++)
        for(int j=2;i*j<=1000000;j++)
            a[i*j]+=i;
}
int main( )
{
    //freopen("1.txt","r",stdin);
    init();
    int s,t,b;
    int kase=0;
    while(cin>>s>>t>>b&&(s||t||b))
    {
        int cnt=0;
        for(int i=s;i<=t;i++)
            if(abs(i-a[i])<=b)
                cnt++;
        printf("Test %d: %d\n",++kase,cnt);
    }
    return 0;
}