解題區/解題區 - UVa

UVa 12425 - Best Friend

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

給一個整數 N，接著詢問 x，有多少 gcd(i, N) <= x。

Sample Input

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5
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8
2
30 3
1
2
10
11 2
1
2

Sample Output

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6
7
Case 1
8
16
28
Case 2
10
10

Solution

先把 N 的因數 f 全部找出，原則上不超過 2000 個，然後建出 phi(N/f[i])。

當要知道 gcd(i, N) = x 時，相當於找到 gcd(i/x, N/x) = 1，任何與 N/x 互質的個數。建造完後，對於每次詢問二分答案即可。

很多地方忘記用 long long，吃了不少 Runtime error。

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#include <stdio.h>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;
#define maxL (1000000>>5)+1
#define GET(x) (mark[x>>5]>>(x&31)&1)
#define SET(x) (mark[x>>5] |= 1<<(x&31))
int mark[maxL];
int P[100000], Pt = 0;
int phi[1048576];
void sieve() {
    register int i, j, k;
    SET(1);
    int n = 1000000;
    for (i = 1; i <= n; i++)
    	phi[i] = i;
    for (i = 2; i <= n; i++) {
        if(!GET(i)) {
            for (k = n/i, j = i*k; k >= i; k--, j -= i)
                SET(j);
            for (j = i; j <= n; j += i)
            	phi[j] = phi[j]/i * (i-1);
            P[Pt++] = i;
        }
    }
}
vector< pair<long long, int> > factor(long long n) {
    long long on = n;
    vector< pair<long long, int> > R;
    
    for(int i = 0, j; i < Pt && (long long)P[i] * P[i] <= n; i++) {
        if(n%P[i] == 0) {		
            for(j = 0; n%P[i] == 0; n /= P[i], j++);
            R.push_back(make_pair((long long)P[i], j));
        }
    }
    if(n != 1)	R.push_back(make_pair(n, 1));
    return R;
}
void make(int idx, int n, long long m, vector< pair<long long, int> > &v, vector<long long> &ret) {
    if(idx == v.size()) {
        ret.push_back(m);
        return;
    }
    long long a = m, b = v[idx].first;
    for(int i = v[idx].second; i >= 0; i--)
        make(idx + 1, n, a, v, ret), a *= b;
}
long long getPhi(long long n) {
    if (n < 1000000)	return phi[n];
    vector< pair<long long, int> > f = factor(n);
    for (int i = 0; i < f.size(); i++)
        n = n / f[i].first * (f[i].first - 1);
    return n;
}
int main() {
    sieve();
    int testcase, cases = 0, Q;
    long long N, X;
    scanf("%d", &testcase);
    while (testcase--) {
        scanf("%lld %d", &N, &Q);
        vector< pair<long long, int> > f = factor(N);
        vector<long long> fa;
        vector<long long> sum;
        make(0, N, 1, f, fa);
        sum.resize(fa.size(), 0);
        sort(fa.begin(), fa.end());
        for (int i = 0; i < fa.size(); i++) {
            if (i)
                sum[i] = sum[i-1] + getPhi(N / fa[i]);
            else
                sum[i] = getPhi(N / fa[i]);
        }
        printf("Case %d\n", ++cases);
        for (int i = 0; i < Q; i++) {
            scanf("%lld", &X);
            if (X <= 0) {
                puts("0");
                continue;
            }
            long long ret = 0;
            int pos = (int)(lower_bound(fa.begin(), fa.end(), X) - fa.begin());
            if (pos >= fa.size() || fa[pos] > X)	pos--;
            ret = sum[pos];
            printf("%lld\n", ret);
        }
    }
    return 0;
}
/*
2
30 3
1
2
10
11 5
0
1
2
11
15
*/