POJ 2773 Happy 2006 (two-point answer + allowance)

Title Link: http://poj.org/problem?id=2773

Test instructions

The number of K and M coprime is obtained;

Analysis:

It is obvious that as the number increases and the number of M coprime is greater, so we can answer two points,

In the middle, we need to use the principle of tolerance to find the number of coprime in [1,mid] and M.

The code is as follows:

#include <iostream> #include <cstring> #include <cstdio> #include <vector>using namespace std;    const int MAXN = 100;typedef long long ll;int m,k,cnt;int p[maxn];void fen (int m)//to M-factor decomposition {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 CalU (ll N)//tolerance count {ll sum=0;        for (ll i=1;i< (1<<cnt); i++) {ll mult=1,f=0;                for (int j=0;j<cnt;j++) {if (i& (1<<j)) {f++;            MULT*=P[J];        }} if (f&1) Sum+=n/mult;    else Sum-=n/mult; } return n-sum;}        int main () {while (~scanf ("%d%d", &m,&k)) {LL l=0,r= ((LL) 1<<62);        LL mid,ans=0;        Fen (m);            while (l<=r) {//two points answer mid= (l+r)/2;            LL sum = CalU (mid);                if (sum>=k) {r=mid-1;            if (sum==k) Ans=mid;        }    else l=mid+1;    } printf ("%i64d\n", ans); } return 0;}

POJ 2773 Happy 2006 (two-point answer + allowance)