数论题集

A - Happy 2006

 POJ - 2773   （欧拉函数）

题意：求第k个与n互质的数（k大于n）

题解：gcd(a,b) = gcd(b,a % b) = gcd(b, a + b * n)可见，与b的gcd == 1的数是可以轮回的，那么这就可以先预先处理出小于n的数厘米的与n互质的，然后开始判断这是在第几个周期里面的第几个数

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<stack>
#include<cstdlib>
#include<queue>
#include<set>
#include<string.h>
#include<vector>
#include<deque>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define eps 1e-4
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl

#define Mod(a,b) a<b?a:a%b+b
typedef long long LL;
typedef long long ll;
const int maxn = 1e6 + 5;

int euler[maxn + 10];

void phi() {
    for (int i = 1; i <= maxn; i++) euler[i] = i;
    for (int i = 2; i <= maxn; i += 2) euler[i] = i / 2;
    for (int i = 3; i <= maxn; i++) {
        if (euler[i] == i) {
            for(int j = i; j <= maxn; j += i)
                euler[j] = euler[j] / i * (i - 1);
        }
    }
}

int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a % b);
}

int main() {
    int n, k;
    phi();
    while (~scanf("%d %d",&n, &k)) {
        int t = k / euler[n];   //第几个周期
        int p = k % euler[n];   //该个周期的第几个
        if (p == 0) {
            t--;
            p = euler[n];
        }
        int i;
        for (i = 1; i <= n; i++) {
            if (gcd(i, n) == 1) p--;
            if (p == 0) break;
        }
        printf("%lld\n",i + (LL) t * n);
    }
}

 

B - Prime Distance

 POJ - 2689　　（素数筛）

题意：给你一个L，U(1<=L< U<=2,147,483,647),The difference between L and U will not exceed 1,000,000，要求在L，U中找一对相邻的素数要求其差值(大减小)最小和最大，并输出

题解：由于L，U过于庞大不可以直接打表，便可采用删去和数的方法，显然L和U如果是和数的话必定还有1e5以下的因子，那就筛取1e5以内的质数，并在L,U中将其倍数删去

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<stack>
#include<cstdlib>
#include<queue>
#include<set>
#include<string.h>
#include<vector>
#include<deque>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define eps 1e-4
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl

#define Mod(a,b) a<b?a:a%b+b
typedef long long LL;
typedef long long ll;
const int maxn = 1e5 + 5;

int prime[maxn + 1];

void getPrime() {
    memset(prime,0,sizeof prime);
    for (int i = 2; i <= maxn; i++) {
        if(!prime[i]) prime[++prime[0]] = i;
        for (int j = 1; j <= prime[0] && prime[j] <= maxn / i; j++) {
            prime[prime[j] * i] = 1;
            if (i % prime[j] == 0) break;
        }
    }
}

int vis[maxn * 10];
int main() {
    getPrime();
//    cout << prime[0] << endl;
    int l, r;
    while (~scanf("%d %d", &l, &r)) {
        memset(vis, 0, sizeof vis);
        if (l == 1) l++;
        for (int i = 1; i < prime[0]; i++) {
            int a = (l - 1) / prime[i] + 1;
            int b = r / prime[i];
            for (int j = a; j <= b; j++)
                if (j > 1) vis[j * prime[i] - l] = 1;
        }
        int p = -1, max_ans = -1, min_ans = inf;
        int x1, x2, y1, y2;
        for (int i = 0; i <= r - l; i++) {
            if (vis[i] == 0) {
                if (p == -1) {
                    p = i;
                    continue;
                }
                if (max_ans < i - p) {
                    max_ans = i - p;
                    x1 = p + l;
                    y1 = i + l;
                }
                if (min_ans > i - p) {
                    min_ans = i - p;
                    x2 = p + l;
                    y2 = i + l;
                }
                p = i;
            }
        }
        if (max_ans == -1)cout << "There are no adjacent primes." << endl;
        else cout << x2 << "," << y2 << " are closest, " << x1 << "," << y1 << " are most distant." << endl;
    }
}