poj 2773 (欧拉函数)

分析： [1,m-1]有phi[m]个数与m互质。则 显然 [n * m + 1, ((n + 1)* m) - 1]也有phi[m]个数与之互质。（n = 0,1,2,...）

RE 的考虑 m = 1 这种情况。

// File Name: 2773.cpp
// Author: Missa_Chen
// Created Time: 2013年05月28日 星期二 13时58分13秒

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<stack>
#include<set>
#include<cstdlib>

using namespace std;

#define LL long long
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 + 10;
int phi[maxn], pn, prim[maxn];
bool p[maxn];
void init_euler(int n)
{
    phi[1] = 0;
    memset(p,0,sizeof(p));
    p[0] = p[1] = 1;
    for (int i = 2; i <= n; ++i)
    {
        if (!p[i])
        {
            phi[i] = i - 1;
            prim[pn++] = i;
        }
        for (int j = 0; j < pn && prim[j] * i <= n; ++j)
        {
            p[prim[j] * i] = 1;
            if (i % prim[j])
                phi[i * prim[j]] = phi[i] * (prim[j] - 1);
            else
            {
                phi[i * prim[j]] = phi[i] * prim[j];
                break;
            }
        }
    }
}
LL gcd(LL a, LL b)
{
    while (b != 0)
    {
        LL c = a % b;
        a = b;
        b = c;
    }
    return a;
    //return b == 0 ? a : gcd(b, a % b);
}
int m, k;

int main()
{
    init_euler((int)1e6);
    while (~scanf("%d%d",&m,&k))
    {
        if (m == 1)
        {
            printf("%d\n",k);
            continue;
        }
        int n = k / phi[m];
        k -= n * phi[m];
        if (k == 0) k = phi[m], n--;
        for (LL i = n * m + 1; ; ++i)
        {
            if (gcd(m, i) == 1LL) k--;
            if (k == 0)
            {
                printf("%lld\n",i);
                break;
            }
        }
    }
    return 0;
}