Happy 2006
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 10309 | Accepted: 3566 |
Description
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.
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
POJ Monthly--2006.03.26,static
题目链接:http://poj.org/problem?id=2773
题目大意:求第k个与m互质的数
题目分析:两种方法,主流方法是分解质因数+二分+容斥,0ms过,先对m分解质因数,然后二分,二分时用容斥计算1到当前数字与m不互质的数的个数,然后用m减,就是与m互质的数的个数。
#include <cstdio> #include <cstring> #define ll long long int const MAX = 1e6 + 6; int fac[MAX]; int n, k, cnt; ll sum; void get_Factor(int x) { cnt = 0; for(int i = 2; i <= x; i++) { if(x % i == 0) fac[cnt ++] = i; while(x % i == 0) x /= i; } if(x > 1) fac[cnt ++] = x; } void DFS(int pos, int num, ll cur, ll x) { if(pos == cnt) { if(cur == 1) return; if(num & 1) sum += x / cur; else sum -= x / cur; return; } DFS(pos + 1, num, cur, x); DFS(pos + 1, num + 1, cur * fac[pos], x); return; } ll cal(ll mid) { sum = 0; DFS(0, 0, 1, mid); return sum; } int main() { while(scanf("%d %d", &n, &k) != EOF) { get_Factor(n); ll l = 0, r = 1e17, mid; while(l <= r) { mid = (l + r) >> 1; if(mid - cal(mid) < k) l = mid + 1; else r = mid - 1; } printf("%lld\n", l); } }
第二种方法利用了欧几里德算法,gcd(a, b) = gcd(b, a % b),令a = km + i,b = m则有:
gcd(km + i, m) = gcd(i, m),说明与m互质的数对m取模有周期性,因此设小于m且与m互质的数有cnt个,第i个为ai,则第m*cnt+i个为m*cnt+ai,其次要注意k整除cnt的时候,取到的实际上是a[cnt]而不是a[0],这个方法跑了2400ms+
#include <cstdio> int const MAX = 1e6 + 5; int a[MAX]; int gcd(int a, int b) { while(b) { int tmp = a; a = b; b = tmp % b; } return a; } int main() { int m, k; while(scanf("%d %d", &m, &k) != EOF) { if(k == 1) { printf("1\n"); continue; } if(m == 1) { printf("%d\n", k); continue; } int cnt = 1; for(int i = 1; i <= m; i++) if(gcd(i, m) == 1) a[cnt ++] = i; cnt --; if(k % cnt == 0) printf("%d\n", (k / cnt - 1) * m + a[cnt]); else printf("%d\n", (k / cnt) * m + a[k % cnt]); } }
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