uva 10837 - A Research Problem(欧拉功能+暴力)

题目链接：uva 10837 - A Research Problem

题目大意：给定一个phin。要求一个最小的n。欧拉函数n等于phin

解题思路：欧拉函数性质有，p为素数的话有phip=p−1;假设p和q互质的话有phip∗q=phip∗phiq

然后依据这种性质，n=pk11(p1−1)∗pk22(p2−1)∗⋯∗pkii(pi−1),将全部的pi处理出来。暴力搜索维护最小值，尽管看上去复杂度很高，可是由于对于垒乘来说，增长很快，所以搜索范围大大被缩小了。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
const int maxp = 1e4;
const int INF = 0x3f3f3f3f;

int ans;
int np, vis[maxp+5], pri[maxp+5];
int nf, fact[maxp+5], v[maxp+5];

void prime_table (int n) {
    np = 0;
    for (int i = 2; i <= n; i++) {
        if (vis[i])
            continue;

        pri[np++] = i;
        for (int j = i * i; j <= n; j += i)
            vis[j] = 1;
    }
}

void get_fact (int n) {
    nf = 0;

    for (int i = 0; i < np && (pri[i]-1) * (pri[i]-1) <= n; i++) {
        if (n % (pri[i]-1) == 0)
            fact[nf++] = pri[i];
    }
}

bool judge (int n) {
    if (n == 2)
        return true;
    for (int i = 0; i < np && pri[i] * pri[i] <= n; i++)
        if (n % pri[i] == 0)
            return false;

    for (int i = 0; i < nf; i++)
        if (v[i] && fact[i] == n)
            return false;
    return true;
}

void dfs (int ret, int cur, int d) {
    if (d == nf) {
        if (judge(cur+1)) {

            if (cur == 1)
                cur = 0;

            ans = min(ans, ret * (cur+1));
        }
        return;
    }

    dfs(ret, cur, d+1);
    if (cur % (fact[d]-1) == 0) {
        v[d] = 1;
        ret *= fact[d];
        cur /= (fact[d]-1);

        while (true) {
            dfs(ret, cur, d+1);

            if (cur % fact[d])
                return;
            ret *= fact[d];
            cur /= fact[d];
        }
        v[d] = 0;
    }
}

int solve (int n) {
    ans = INF;
    get_fact(n);
    memset(v, 0, sizeof(v));
    dfs(1, n, 0);
    return ans;
}

int main () {
    prime_table(maxp);
    int cas = 1, n;
    while (scanf("%d", &n) == 1 && n) {
        printf("Case %d: %d %d\n", cas++, n, solve(n));
    }
    return 0;
}

版权声明：本文博客原创文章。博客，未经同意，不得转载。