数论

 

Quite Good Numbers

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB

Total submit users: 77, Accepted users: 57

Problem 12876 : No special judgement

Problem 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 2
2 100 0
1000 9999 3
0 0 0

Sample Output

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

Problem Source

HNU Contest 

---

Mean:

让你求从sta到end这个区间中有多少个数满足：abs(sum-i)<=bad。其中sum是i所有的因子之和，bad是给定的值，代表误差。

 

analyse:

由于数字很大，必须打表，我们将10^6次方内i的因子之和求出来。

从筛法求素数得到的启发，方法很巧妙，具体看代码。

Time complexity：O(n)

Source code：

//Memory   Time
// 4988K    40MS
// by : Snarl_jsb
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<iomanip>
#include<string>
#include<climits>
#include<cmath>
#define MAX 1000005
#define LL long long
using namespace std;
int sta,stop,bad,ans,kase=1;
int sum[MAX];
void make_table()
{
    for(int i=2;i<=MAX;i++)
    {
        sum[i]++;
        for(int j=2;i*j<=MAX;j++) sum[i*j]+=i;
    }
}
int main()
{
    make_table();
    while(scanf("%d %d %d",&sta,&stop,&bad),ans=0,sta+stop+bad)
    {
        printf("Test %d: ",kase++);
        for(int i=sta;i<=stop;i++)
        {
            if(abs(sum[i]-i)<=bad) ans++;
        }
        cout<<ans<<endl;
    }
    return 0;
}