POJ 2773 Happy 2006 (容斥原理)
题目是给出m,k。找到跟第k个跟m互素的数是多少。
构造肯定不行,再加上数据范围,只能二分。思路是二分枚举[1,2^64]范围内所有的数x,找到1到x范围内与m不互素的数的个数y(用容斥原理)。然后用x - y,如果等于k就是结果。
找到1到x范围内与m不互素的数的个数y:这个过程可以先把m分解质因子,记录m所有的质因子。f[i]表示含有i个质因子的数的个数。ans = m - f(1) + f(2) - f(3) ....
ps:这里二分要找满足 == k最左边的数,推了半天发现把二分写错了。。。T_T
//#pragma comment(linker,"/STACK:327680000,327680000") #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <algorithm> #include <string> #include <set> #include <functional> #include <numeric> #include <sstream> #include <stack> #include <map> #include <queue> #define CL(arr, val) memset(arr, val, sizeof(arr)) #define REP(i, n) for((i) = 0; (i) < (n); ++(i)) #define FOR(i, l, h) for((i) = (l); (i) <= (h); ++(i)) #define FORD(i, h, l) for((i) = (h); (i) >= (l); --(i)) #define L(x) (x) << 1 #define R(x) (x) << 1 | 1 #define MID(l, r) (l + r) >> 1 #define Min(x, y) (x) < (y) ? (x) : (y) #define Max(x, y) (x) < (y) ? (y) : (x) #define E(x) (1 << (x)) #define iabs(x) (x) < 0 ? -(x) : (x) #define OUT(x) printf("%I64d\n", x) #define Read() freopen("data.in", "r", stdin) #define Write() freopen("data.out", "w", stdout); typedef long long LL; const double eps = 1e-8; const double pi = acos(-1.0); const double inf = ~0u>>2; using namespace std; const int N = 1000010; int p[N], cnt; void init(int m) { cnt = 0; for(int i = 2; i*i <= m; ++i) { if(m%i == 0) { p[cnt++] = i; while(m%i == 0) {m /= i;} } } if(m != 1) p[cnt++] = m; } LL cal(LL n) { int i, j, bit; LL res = 0, sum; for(i = 1; i < 1<<cnt; ++i) { bit = 0; sum = 1; for(j = 0; j < cnt; ++j) { if(i&(1<<j)) { bit++; sum *= p[j]; } } if(bit&1) res -= n/sum; else res += n/sum; } return n + res; } int main() { //Read(); int m, k; while(cin >> m >> k, !cin.eof()) { init(m); LL l = 1, r = (1LL<<60), mid, tmp; while(r - l > 0) { mid = MID(l, r); tmp = cal(mid); if(tmp >= k) r = mid; else l = mid + 1; } printf("%lld\n", l); } return 0; }