HDU 4135 Co-prime (分解质因数 ＋ 容斥)...
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其他好文 时间：
2015-08-16 10:48:44
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97
Co-primeTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2307Accepted Submission(s): 861 Problem D...
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其他好文 时间：
2015-08-16 07:09:15
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161
Distinct primes factors
Problem 47
The first two consecutive numbers to have two distinct prime factors are:14 = 2 × 7
15 = 3 × 5The first three consecutive numbers to have three distinct prime fact...
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其他好文 时间：
2015-08-15 21:36:32
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233
看到了一个很神的模板，具体什么意思还不理解，先马了再研究~ 1 bool prime (long long num) 2 { 3 if (num == 2 || num == 3 || num == 5) 4 return true; 5 if (nu...
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其他好文 时间：
2015-08-15 21:19:57
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108
DescriptionA number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18,...
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其他好文 时间：
2015-08-15 10:11:34
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161
#include#include#includeusing namespace std;int prime (int a){ bool pr; pr=false; int s; s=0; for (int i=1;i<=a;i++) if ...
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编程语言 时间：
2015-08-14 18:51:58
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156
Description
If a and d are relatively prime positive integers, the arithmetic sequence beginning with a and increasing by d, i.e., a, a + d, a + 2d, a + 3d, a +
4d, ..., contains infinitely man...
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其他好文 时间：
2015-08-14 17:07:58
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77
Problem Description
A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime...
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其他好文 时间：
2015-08-14 15:45:56
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113
下面是埃拉托斯特尼筛法的实现代码：
boolean[] sieveOfEratosthenes(int max)
{
boolean[] flags = new boolean[max + 1];
int count = 0;
init(flags);//将flags中0,1元素除外的所有元素设为true
int prime = 2;
while (prime
{
//划掉余下...
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其他好文 时间：
2015-08-13 22:13:15
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175
1059. Prime Factors (25)Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1^k1* p2^k2...
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其他好文 时间：
2015-08-12 23:16:27
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130
