Home_W的超级数学题(第k个和m互素的数

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Home_W的超级数学题

TimeLimit:3000MS  MemoryLimit:65536K

64-bit integer IO format:%lld

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Problem Description

Home_W看到T^T竟然也出了一道跟他差不多短的题目，于是他嘴角微微一笑，拿出了他珍藏多年的超级数学题。
给定一个数m和一个数k，问第k个和m互素的数是多少?（从小到大）

Input

多组测试数据. 每行包含两个数，分别是m (1 <= m <= 1000000), K (1 <= K <= 100000000).

Output

输出第k个和m互素的数。

SampleInput

2006 1
2006 2
2006 3

SampleOutput

1
3
5

Submit

题意： 

求第k个和m互素的数。m<=106,k<=108 
分析：
如果gcd(a,m)=1,那么gcd(a+k∗m,m)=1,k∈Z

证明:只需要证明gcd(a+m,m)=1.令d=gcd(a+m,m)，设a+m=d∗p1,m=d∗p2且p1>p2,,移项可得:a=d∗(p1−p2),如果d!=1，那么gcd(a,m)!=1，与原条件不符，故gcd(a,m)=1，同理可证gcd(a+k∗m,m)=1,k∈Z.

所以先求个欧拉函数，再暴力搞一下就好了。

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=2e3+10;
const int maxx=1e6+100;
const double EPS=1e-8;
const double eps=1e-8;
//const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

LL oula(LL n)
{
LL ans=n;
if(n%2==0)//只搜n/2次，不然可能超一组 （也可能不超）
{
while(n%2==0) n/=2;
ans=ans/2;
}
for(long long i=3;i*i<=n;i+=2)
//每找到一个就更改上界，这个优化很多，所以直接copy刘汝佳的代码会超时
{
if(n%i==0)
{
while(n%i==0) n/=i;
ans=ans/i*(i-1);
}
}
if(n>1) ans=ans/n*(n-1);
return ans;
}
int m,k;
int main()
{
W(scan_d(m),scan_d(k))
{
if(m==1)
{
print(k);
continue;
}
LL num=oula(m);
int t=k/num;
k%=num;
if(k==0)
{
k=num;
t--;
}
int st = 0;
W(k--)
{
W(__gcd(st, m) != 1) st++;
st++;
}
print((LL)t * m + st - 1);
}
}