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. |
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: |
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; }