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;
}