hdu

Problem Description

Please write a program to calculate the k-th positive integer that is coprime with m and n simultaneously. A is coprime with B when their greatest common divisor is 1.

Input

The first line contains one integer T representing the number of test cases.
For each case, there’s one line containing three integers m, n and k (0 < m, n, k <= 10^9).

Output

For each test case, in one line print the case number and the k-th positive integer that is coprime with m and n.
Please follow the format of the sample output.

Sample Input

3
6 9 1
6 9 2
6 9 3

Sample Output

Case 1: 1
Case 2: 5
Case 3: 7

Author

xay@whu

Source

The 5th Guangting Cup Central China Invitational Programming Contest

题解

• 若要一个数x与m，n均互质，x则不能含义m，n的素因子。因为k很大，暴力求会超时。
• 由容斥原理我们知道，只要x里没有m,n的素因子就是互质，那么1-n内的，对m，n互质的数的个数我们就可以认为是n-cal(n)，（cal()函数为计算1-n中与m,n不互质数的个数）
• 对于m，n的素因子，我们不能通过m*n求，而时分别求m,n,然后排序得到去除重复的素因子数组fac[]。
• 二分搜索，那么对于第k个互质数，我们只要搜索n-cal(n)=k时，说明，1-n区间有k个互质数，但n不一定是第k个互质数，这时候要对n递减，只要n不满足这个条件我们就可以返回n+1.

#include <cstdio>#include <iostream>#include <algorithm>#include <cmath>using namespace std;typedef long long LL;const LL INF=1e18+5;int fac[1005];//数组大小视具体情况而定int cnt;void getFactor(LL n){//求n的所有质因素    LL m=sqrt(n+0.5);    for(LL i=2;i<=m&&n;i++){        if(n%i==0){            fac[cnt++]=i;            while(n&&n%i==0) n/=i;        }    }    if(n>1) fac[cnt++]=n;}LL cal(LL n){//容斥    LL res=0;    for(LL i=1;i<(1LL<<cnt);i++){        LL mult=1,ones=0;        for(int j=0;j<cnt;j++){            if(i&(1<<j)){                ones++;                mult*=fac[j];            }        }        if(ones&1) res+=n/mult;        else res-=n/mult;    }    return res;}LL bin_serach(LL k){//二分搜索    LL l=0,r=INF,mid;    while(l<r){        mid=(l+r)>>1;        LL d=mid-cal(mid);        if(d>=k) r=mid;        else l=mid+1;    }    return r;}int main(){    int T;    scanf("%d",&T);    for(int cas=1;cas<=T;++cas){        LL n,m,k;        cnt=0;        scanf("%lld%lld%lld",&n,&m,&k);        getFactor(n);        getFactor(m);        sort(fac,fac+cnt);        cnt=unique(fac,fac+cnt)-fac;//去重        printf("Case %d: %lld\n",cas,bin_serach(k));    }    return 0;}

注：这个是二分搜索的特殊形式，实在不懂就用下面这个。
二分搜索详解 戳这<<

LL bin_serach(LL k){//二分搜索    LL l=0,r=INF,mid;    while(l<=r){        mid=(l+r)>>1;        LL d=mid-cal(mid);        if(d>k) r=mid-1;        else if(d<k) l=mid+1;        else{            while(d==k){                mid--;                d=mid-cal(mid);            }            return mid+1;        }    }    return r;}