hdu 1695 GCD (欧拉函数、容斥原理)
2015-08-06 19:21
519 查看
GCD
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7357 Accepted Submission(s): 2698
Problem Description
Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the total number of different number
pairs.
Please notice that, (x=5, y=7) and (x=7, y=5) are considered to be the same.
Yoiu can assume that a = c = 1 in all test cases.
Input
The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 3,000 cases.
Each case contains five integers: a, b, c, d, k, 0 < a <= b <= 100,000, 0 < c <= d <= 100,000, 0 <= k <= 100,000, as described above.
Output
For each test case, print the number of choices. Use the format in the example.
Sample Input
2 1 3 1 5 1 1 11014 1 14409 9
Sample Output
Case 1: 9 Case 2: 736427 HintFor the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).
Source
2008 “Sunline Cup” National Invitational Contest
题目大意:求出[a,b]和[c,d]区间里面gcd(x,y)=k的数的对数。
思路:既然是求gcd为k的数的对数,不妨先将b和d都除以k,这样问题就转化为[1,n]和[1,m]区间里面gcd(x,y)为1 的数的对数。因为题目里已经说明a和c 可以认为是1,这样就更简单了。对于一个[1,n]的区间,我们可以用欧拉函数算出总对数。
那么问题就可以分解成2个:
1、在[1,n]上用欧拉函数算出总对数。
2、在[n+1,m]上,计算在[1,n]里面的总对数,可以用容斥原理。
#include<stdio.h> #include<math.h> #include<string.h> #include<iostream> #include<algorithm> #include<vector> #define min(a,b) a<b?a:b #define max(a,b) a>b?a:b #define Max 100005 #define LL __int64 using namespace std; LL sum[Max],tot; int p[Max][20]; int num[Max]; void init() { sum[1]=1; for(int i=2;i<Max;i++) sum[i]=i; for(int i=2;i<Max;i++) if(sum[i]==i) for(int j=i;j<Max;j+=i) sum[j]=sum[j]/i*(i-1); } void init2() { LL x,k,i,j; for( i=1;i<=Max;i++) { x=i;k=0; for(j=2;j<=sqrt(i);j++) { if(x%j==0){ while(x%j==0)x=x/j; // p[i].push_back(j); p[i][num[i]++]=j; } } if(x>1)p[i][num[i]++]=x; } } LL dfs(int n,int b,int x,int k) { LL ans=0; for(int i=x;i<k;i++) { ans+=b/p [i]-dfs(n,b/p [i],i+1,k); } return ans; } int main() { LL T,a,b,c,d,k; int i,j,t; init(); init2(); // printf("%I64d %I64d\n",sum[2],sum[3]); scanf("%I64d",&T); t=0; while(T--) { tot=0; t++; scanf("%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&k); printf("Case %d: ",t); if(k==0){printf("0\n");continue;} b=b/k; d=d/k; int m; m=min(b,d); d=max(b,d); b=m; for(i=1;i<=b;i++) tot=tot+sum[i]; for(i=b+1;i<=d;i++) { // printf("%d\n",p[i].size()); tot+=b-dfs(i,b,0,num[i]); } printf("%I64d\n",tot); } return 0; }
相关文章推荐
- hdu 1372
- 开通博客之初衷
- [转]WIN7系统安装Apache 提示msvcr110.DLL
- memcached源码分析之线程池机制(一)
- centos yum安装mysql
- 如何长时间高效学习?(2)
- 东软毕设
- hdoj 2717 Catch That Cow
- zoj 1671 Walking Ant
- pyqt 用py2exe打包出错的解决办法
- 偶然间做起题来--仅为纪念
- Python中的类属性和实例属性以及静态方法和类方法
- string的erase方法
- iOS程序启动的时候隐藏状态栏,启动完成之后显示状态栏
- RTX与SVN使用手册适用于新手
- xhprof(PHP性能分析工具)
- hdu 5360 Hiking(2015 Multi-University Training Contest 6)
- 经典排序算法之简单选择排序
- 93. Restore IP Addresses
- 七把利器助你打造充满竞争力的商业模式