| 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 badnessA 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 |
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#include #include #include #include #include #include #include #include #include #include #include #define MAX 1000005#define LL long longusing 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< <