http://acm.hdu.edu.cn/showproblem.php?pid=3388

容斥定理，题意：给两个数n，m，问第k个跟他们互质的数是啥，刚看到这一题的时候，果断弄出n，m的lcm，但是超时了，后来左改右改连滚带爬过了

Coprime

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 760    Accepted Submission(s): 201

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

36 9 16 9 26 9 3

 

Sample Output

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

 

Author

xay@whu

 

#include <cstdio>#include <vector>#include <algorithm>using namespace std;vector<__int64> fac;void get_fac(__int64 n,__int64 m){    for(__int64 i=2;i*i<=n;i++)        if(n%i==0){            fac.push_back(i);            while(n%i==0)n/=i;        }    if(n>1)        fac.push_back(n);    for(__int64 i=2;i*i<=m;i++)        if(m%i==0){            fac.push_back(i);            while(m%i==0)m/=i;        }    if(m>1)fac.push_back(m);    sort(fac.begin(),fac.end());    fac.erase(unique(fac.begin(),fac.end()),fac.end());}__int64 gcd(__int64 x,__int64 y){    return y?gcd(y,x%y):x;}__int64 fun(__int64 x){    __int64 sum=0;    __int64 cnt=fac.size();    for(int num=1;num<(1<<cnt);num++){//cnt个集合，容斥定理中就有2的cnt次方-1项,num是二进制压缩//3为11，代表A1交A2        __int64 mult=1,ones=0;        for(int i=0;i<cnt;i++){//i表示第几个集合，例如i=2，表示为100，表示为第三个集合            if(num&(1<<i)){//当i为0的时候，判断第一个集合是否进行交集操作            ones++;//当ones为偶数的时候，那么就是减，当noes为奇数的时候为加            mult*=fac[i];            }        }    if(ones%2) sum+=x/mult;    else sum-=x/mult;    }    return x-sum;}//这里的fac是指一个数n的素因子，a表示1~a大小的集合，所求是1~a内与n互质的数的个数__int64 get_ans(__int64 low,__int64 high,__int64 n,__int64 m,__int64 k){//注意这里二分的次序，如果次序颠倒了，会有意想不到的事情，一开始没注意，超时无数    __int64 mid=(low+high)/2;    if (fun(mid)>k) return  get_ans(low,mid-1,n,m,k);    else if(fun(mid)<k)return get_ans(mid+1,high,n,m,k);    else {        for(;;mid--){            if(gcd(mid,n)==1&&gcd(mid,m)==1)    {                return mid;            }        }    }}int main(){    __int64 m,n,k,t;    scanf("%I64d",&t);    __int64 maxn=(__int64)1<<62;    for(int Case=1;Case<=t;Case++){        scanf("%I64d%I64d%I64d",&m,&n,&k);        fac.clear();        get_fac(n,m);        __int64 ans=get_ans(1,maxn,n,m,k);        printf("Case %d: %I64d\n",Case,ans);    }    return 0;}