POJ 2773 Happy 2006
Happy 2006
Time Limit: 3000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 277364-bit integer IO format: %lld Java class name: Main
Two positive integers are said to be relatively prime to each other if the Great Common Divisor (GCD) is 1. For instance, 1, 3, 5, 7, 9...are all relatively prime to 2006.
Now your job is easy: for the given integer m, find the K-th element which is relatively prime to m when these elements are sorted in ascending order.
Now your job is easy: for the given integer m, find the K-th element which is relatively prime to m when these elements are sorted in ascending order.
Input
The input contains multiple test cases. For each test case, it contains two integers m (1 <= m <= 1000000), K (1 <= K <= 100000000).
Output
Output the K-th element in a single line.
Sample Input
2006 1 2006 2 2006 3
Sample Output
1 3 5
Source
解题:容斥原理
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 typedef long long LL; 6 const LL INF = 0x3f3f3f3f3f3f3f3f; 7 const int maxn = 50; 8 int p[maxn],tot; 9 void init(int x){ 10 tot = 0; 11 for(int i = 2; i*i <= x; ++i){ 12 if(x%i == 0){ 13 p[tot++] = i; 14 while(x%i == 0) x /= i; 15 } 16 } 17 if(x > 1) p[tot++] = x; 18 } 19 LL solve(LL x){ 20 LL ret = 0; 21 for(int i = 1; i < (1<<tot); ++i){ 22 int cnt = 0,tmp = 1; 23 for(int j = 0; j < tot; ++j){ 24 if((i>>j)&1){ 25 ++cnt; 26 tmp *= p[j]; 27 } 28 } 29 if(cnt&1) ret -= x/tmp; 30 else ret += x/tmp; 31 } 32 return x + ret; 33 } 34 int main(){ 35 int n,k; 36 while(~scanf("%d%d",&n,&k)){ 37 LL ret = 0,low = 1,high = INF; 38 init(n); 39 while(low <= high){ 40 LL mid = (low + high)>>1; 41 if(solve(mid) >= k){ 42 ret = mid; 43 high = mid - 1; 44 }else low = mid + 1; 45 } 46 printf("%I64d\n",ret); 47 } 48 return 0; 49 }
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