poj 2773Happy 2006

Happy 2006

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 10620 Accepted: 3709

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 12006 22006 3

Sample Output

135

Source

POJ Monthly--2006.03.26,static

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思路：

     就是在0到0x3f3f3f3f3f3f3f3f(无穷大的数)之间找一个数，它是与n互质的数的第k大数！（但是不知道二分查找的时候为什么找到了还要循环！）

代码：

#include <stdio.h>#include <string.h>#include <algorithm>#define LL long longusing namespace std;LL p[10]={0},k;void getp(int n){k=0;for(int i=2;i*i<=n;i++){if(n%i==0)p[k++]=i;while(n%i==0)n/=i;}if(n>1) p[k++]=n;}LL nop(LL m){LL que[10005],top=0;que[top++]=-1;for(LL i=0;i<k;i++){LL t=top;for(LL j=0;j<t;j++){que[top++]=que[j]*p[i]*(-1);}}LL sum=0;for(int i=1;i<top;i++){sum+=m/que[i];}return sum;}int main(){int m,kk;while(scanf("%d%d",&m,&kk)!=EOF){getp(m);LL l=0,mid,r=0x3f3f3f3f3f3f3f3f;LL ans;while(l<=r){mid=(l+r)>>1;LL t=mid-nop(mid);//这个t一定是LL型的，不能是int型的！ if(t<kk){l=mid+1;}else if(t==kk){r=mid-1;//之前没写这个就不对，不知道为什么？ ans=mid;//为什么找到之后还要进行循环？//break; //而不能break； }else{r=mid-1;}}printf("%lld\n",ans);}return 0;}