程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> POJ2773---Happy 2006(容斥+二分)

POJ2773---Happy 2006(容斥+二分)

編輯：C++入門知識

POJ2773---Happy 2006(容斥+二分)

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

二分區間上限mid,計算[1, mid]裡與n互質的數個數,多次二分以後,mid的值就是第k個和n互質的數

/*************************************************************************
    > File Name: POJ2773.cpp
    > Author: ALex
    > Mail: [email protected] 
    > Created Time: 2015年05月28日 星期四 19時36分15秒
 ************************************************************************/

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair  PLL;

vector  mst;

LL calc(LL high) {
    int size = mst.size();
    LL ans = 0;
    for (int i = 1; i < (1 << size); ++i) {
        int bits = 0;
        LL P = 1;
        for (int j = 0; j < size; ++j) {
            if (i & (1 << j)) {
                ++bits;
                P *= mst[j];
            }
        }
        if (bits & 1) {
            ans += high / P;
        }
        else {
            ans -= high / P;
        }
    }
    return high - ans;
}

int main() {
    LL m, k;
    while (cin >> m >> k) {
        mst.clear();
        LL tmp = m;
        for (int i = 2; i * i <= tmp; ++i) {
            if (tmp % i == 0) {
                mst.push_back(i);
                while (tmp % i == 0) {
                    tmp /= i;
                }
            }
        }
        if (tmp > 1) {
            mst.push_back(tmp);
        }
        LL ans = -1, mid;
        LL l = 1, r = (1LL << 60);
        while (l <= r) {
            mid = (l + r) >> 1;
            LL tmp = calc(mid);
            if (tmp >= k) {
                r = mid - 1;
                ans = mid;
            }
            else {
                l = mid + 1;
            }
        }
        printf("%lld\n", ans);
    }
}