poj2773 Happy 2006(二分+容斥)

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题目链接：点这里！！！！

题意：

给你两个整数m(1<=m<=1e6),k(1<=k<=1e8)。求第k个与m互质的数是多少。

题解：

直接二分+容斥。

代码：

#include<cstdio>#include<cstring>#include<iostream>#include<sstream>#include<algorithm>#include<vector>#include<bitset>#include<set>#include<queue>#include<stack>#include<map>#include<cstdlib>#include<cmath>#define PI 2*asin(1.0)#define LL long long#define pb push_back#define pa pair<int,int>#define clr(a,b) memset(a,b,sizeof(a))#define lson lr<<1,l,mid#define rson lr<<1|1,mid+1,r#define bug(x) printf("%d++++++++++++++++++++%d\n",x,x)#define key_value ch[ch[root][1]][0]C:\Program Files\Git\binconst LL  MOD = 1E9+7;const LL N = 1e3+15;const int maxn = 5e5+15;const int letter = 130;const LL INF = 1e18;const double pi=acos(-1.0);const double eps=1e-10;using namespace std;inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}int x,k;int vis[N],prime[N],cnt=0;int ps[30],num;void init(){    vis[1]=1;    for(int i=2;i<N;i++){        if(!vis[i]) prime[++cnt]=i;        for(int j=1;j<=cnt&&i*prime[j]<N;j++){            vis[i*prime[j]]=1;            if(i%prime[j]==0) break;        }    }}LL check(LL x){    ///[1,x]    LL sum=x,ans=1,res=0;    int cc=0;    for(int i=1;i<(1<<num);i++){        cc=0;        LL ans=1;        for(int j=0;j<num;j++){            if(i&(1<<j)){                cc++;                ans*=1ll*ps[j];            }        }        if(cc%2) res+=x/ans;        else res-=x/ans;    }    return sum-res;}int main(){    init();    while(scanf("%d%d",&x,&k)!=EOF){        num=0;        for(int i=1;i<=cnt;i++){            if(x%prime[i]==0){                while(x%prime[i]==0) x/=prime[i];                ps[num++]=prime[i];            }        }        if(x!=1) ps[num++]=x;        LL l=1,r=INF,mid;        LL flag;        while(l<r){            mid=(l+r)/2;            flag=check(mid);            if(flag<k) l=mid+1;            else r=mid;        }        printf("%I64d\n",r);    }    return 0;}/*12 4*/

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