Co-prime
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1412 Accepted Submission(s): 531Problem DescriptionGiven a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.
Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.InputThe first line on input contains T (0 < T <= 100) the number of test cases, each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 1015) and (1 <=N <= 109).OutputFor each test case, print the number of integers between A and B inclusive which are relatively prime to N. Follow the output format below.Sample Input2 1 10 2 3 15 5Sample OutputCase #1: 5 Case #2: 10HintIn the first test case, the five integers in range [1,10] which are relatively prime to 2 are {1,3,5,7,9}.
题意：求区间[a, b]内与ｎ互质的数的个数。
思路：如果某个数与ｎ互质，那么这个数一定和ｎ没有公共因子。所以题目就转化为有多少个数与ｎ无公共因子。
可以通过求区间内有多少个数与ｎ存在公共因子来得到答案。筛去２的倍数，３的倍数，５的倍数。。。容斥就可以啦。
Accepted Code:
1 /*************************************************************************
2 > File Name: 4135.c
3 > Author: Stomach_ache
4 > Mail: sudaweitong@gmail.com
5 > Created Time: 2014年09月05日 星期五 16时33分45秒
6 > Propose:
7 ************************************************************************/
8 #include <cstdio>
9 #include <vector>
10 #include <cstring>
11 #include <cstdlib>
12 #include <iostream>
13 using namespace std;
14 /*Let's fight!!!*/
15
16 typedef long long LL;
17
18 LL gcd(LL a, LL b) {
19 if (!b) return a;
20 return gcd(b, a % b);
21 }
22
23 LL cal(const vector<int> &var, const LL &n) {
24 LL sz = var.size(), res = 0;
25 for (LL i = 1; i < (1<<sz); i++) {
26 int num = 0;
27 for (LL j = i; j != 0; j >>= 1) if (j & 1) num++;
28 LL lcm = 1;
29 for (LL j = 0; j < sz; j++) {
30 if ((i >> j) & 1) lcm = lcm / gcd(lcm, var[j]) * var[j];
31 if (lcm > n) break;
32 }
33 if (num % 2 == 0) res -= n / lcm;
34 else res += n / lcm;
35 }
36
37 return res;
38 }
39
40 int main(void) {
41 ios::sync_with_stdio(false);
42 int T, cas = 1;
43 cin >> T;
44 while (T--) {
45 LL a, b, n, x;
46 cin >> a >> b >> n;
47
48 vector<int> var;
49 x = n;
50 for (int i = 2; i * i <= x; i++) {
51 if (x % i == 0) {
52 var.push_back(i);
53 while (x % i == 0) x /= i;
54 }
55 }
56 if (x > 1) var.push_back(x);
57
58 LL res = b - a + 1 - cal(var, b) + cal(var, a - 1);
59 cout << "Case #" << cas++ << ": " << res << endl;
60 }
61
62 return 0;
63 }
How many integers can you find
Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4205 Accepted Submission(s): 1198Problem DescriptionNow you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,10}, all the integers of the set can be divided exactly by 2 or 3. As a result, you just output the number 7.InputThere are a lot of cases. For each case, the first line contains two integers N and M. The follow line contains the M integers, and all of them are different from each other. 0<N<2^31,0<M<=10, and the M integer are non-negative and won’t exceed 20.OutputFor each case, output the number.Sample Input12 2 2 3Sample Output7
和上题一样。。。
Accepted Code:
/*************************************************************************
> File Name: 1796_dfs.cpp
> Author: Stomach_ache
> Mail: sudaweitong@gmail.com
> Created Time: 2014年09月06日 星期六 08时28分01秒
> Propose:
************************************************************************/
#include <cmath>
#include <string>
#include <cstdio>
#include <fstream>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
/*Let's fight!!!*/
typedef long long LL;
int a[15], n, m;
LL gcd(LL a, LL b) {
if (!b) return a;
return gcd(b, a % b);
}
void dfs(LL now, int num, LL lcm, LL &res) {
lcm = lcm / gcd(lcm, a[now]) * a[now];
if (num % 2 == 0) res -= n / lcm;
else res += n / lcm;
for (int i = now + 1; i < m; i++)
dfs(i, num + 1, lcm, res);
}
int main(void) {
ios::sync_with_stdio(false);
while (cin >> n >> m) {
int cnt = 0;
for (int i = 0; i < m; i++) {
int x;
cin >> x;
if (x > 0) a[cnt++] = x;
}
m = cnt;
LL res = 0;
n--;
for (int i = 0; i < m; i++) {
dfs(i, 1, a[i], res);
}
cout << res << endl;
}
}
//位运算实现
/*************************************************************************
> File Name: 1796.cpp
> Author: Stomach_ache
> Mail: sudaweitong@gmail.com
> Created Time: 2014年09月05日 星期五 21时32分48秒
> Propose:
************************************************************************/
#include <cmath>
#include <string>
#include <cstdio>
#include <vector>
#include <fstream>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
/*Let's fight!!!*/
typedef long long LL;
int a[12];
LL gcd(LL a, LL b) {
if (!b) return a;
return gcd(b, a % b);
}
int main(void) {
ios::sync_with_stdio(false);
LL n, m;
while (cin >> n >> m) {
int d = 0;
for (int i = 0; i < m; i++) {
int x;
cin >> x;
if (x > 0 && x <= n) a[d++] = x;
}
n--;
LL res = 0;
for (LL i = 1; i < (1 << d); i++) {
int num = 0;
for (LL j = i; j != 0; j >>= 1) num += j & 1;
LL lcm = 1;
for (LL j = 0; j < d; j++) {
if ((i >> j) & 1) {
lcm = lcm / gcd(lcm, a[j]) * a[j];
if (lcm > n) break;
}
}
if (num % 2 == 0) res -= n / lcm;
else res += n / lcm;
}
cout << res << endl;
}
return 0;
}