HDU4135——Co-prime(数论,容斥原理)
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题目:
Co-prime
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3490 Accepted Submission(s): 1379
Problem Description
Given 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.
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.
Input
The 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).
Output
For 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 Input
21 10 23 15 5
Sample Output
Case #1: 5Case #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}.
Source
The Third Lebanese Collegiate Programming Contest
题意:求区间[a,b]上与p互质的数字的个数,其中(1 <= a <=b <= 1015) and (1 <=p <= 109)
思路:首先直接求a,b上和p互质的数并不方便。所以考虑分别求[1,a-1],和[1,b]上和p互质的数的个数。它们的差就是答案。
那么如何去求[1,n]上与p互质的数的个数?
这时候就需要用到容斥原理了。以p有三个不同的的质因数为例。设为a,b,c。
那么[1,n]上不和p互质的数的个数为,m=n/a+n/b+n/c-n/(a*b)-n/(a*c)-n/(b*c)+n/(a*b*c) (奇加偶减)
互质的数的个数就是n-m
计算的时候可以用dfs,队列数组,还有状态压缩。
#include <cmath>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <algorithm>#include <string>#include <iostream>using namespace std;#define MAXN 55#define INF 1e9+7#define MODE 1000000typedef long long ll;int t;ll a,b,n;vector <int>ans;ll solve(ll n){ //que用来保存每一项分母,分母是偶数个相乘为负,奇数个为正 ll que[10000]; int t=0; que[t++]=-1; for(int i=0;i<ans.size();i++) { int k=t; for(int j=0;j<k;j++) { que[t++]=que[j]*ans[i]*(-1); } } ll sum=0; for(int i=1;i<t;i++) sum=sum+n/que[i]; return n-sum;}int main(){ scanf("%d",&t); for(int cas=1;cas<=t;cas++) { scanf("%I64d%I64d%I64d",&a,&b,&n); ans.clear(); int temp=n; for(int i=2;i*i<=temp;i++) { if(temp%i==0){ ans.push_back(i); while(temp%i==0) temp/=i; } } if(temp>1) ans.push_back(temp); ll s1=solve(a-1); ll s2=solve(b); printf("Case #%d: %I64d\n",cas,s2-s1); }}
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