2018.11.6刷题记录

数论基本糙作：

　　gcd，快速幂，逆元，欧拉函数，分解因数balabala一通乱搞。

 

POJ1845 Sumdiv （数论：算数基本定理+数论基本操作）

题目：

 

Sumdiv
Time Limit: 1000MS        Memory Limit: 30000K
Total Submissions: 29012        Accepted: 7127

Description
Consider two natural numbers A and B. Let S be the sum of all natural divisors of A^B. Determine S modulo 9901 (the rest of the division of S by 9901).

Input
The only line contains the two natural numbers A and B, (0 <= A,B <= 50000000)separated by blanks.

Output
The only line of the output will contain S modulo 9901.

Sample Input

2 3

Sample Output

15

Hint
2^3 = 8.
The natural divisors of 8 are: 1,2,4,8. Their sum is 15.
15 modulo 9901 is 15 (that should be output). 

 

代码：

 

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
typedef long long ll;
const int MAX_N = 10000;
const int MOD = 9901;

ll fpow(ll a, ll p) {
    ll ans = 1;
    while (p) {
        if (p&1) ans = (ans*a) % MOD;
        a = (a*a) % MOD;
        p >>= 1;
    }
    return ans;
}

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

ll m;
ll p[100], c[100];
ll getFactors(ll x)
{
    m = 0;
    for (int i = 1; prime[i] <= x/prime[i]; i++) {
        if (x%prime[i] == 0) {
            p[m] = prime[i];
            c[m] = 0;
            while (x%prime[i] == 0) {
                c[m]++;
                x /= prime[i];
            }
            m++;
        }
    }
    if (x > 1) {
        p[m] = x;
        c[m] = 1;
        m++;
    }
    return m;
}

int main()
{
    getPrime();
    ll A, B;
    cin >> A >> B;
    getFactors(A);
    ll ans = 1;
    for (int i = 0; i < m; i++) {
        if (p[i] % MOD == 1) {
            ans = (B*c[i]+1)%MOD * ans % MOD;
            continue;
        }
        ll a = (fpow(p[i], B*c[i]+1)-1+MOD) % MOD;
        ll b = fpow(p[i]-1, MOD-2);
        ans = ans * a % MOD * b % MOD;
    }
    cout << ans << endl;
    return 0;
}

 

POJ3696 The Luckiest Number （数论：欧拉定理+数论基本操作）

题目：

The Luckiest number
Time Limit: 1000MS        Memory Limit: 65536K
Total Submissions: 6829        Accepted: 1820

Description

Chinese people think of '8' as the lucky digit. Bob also likes digit '8'. Moreover, Bob has his own lucky number L. Now he wants to construct his luckiest number which is the minimum among all positive integers that are a multiple of L and consist of only digit '8'.

Input

The input consists of multiple test cases. Each test case contains exactly one line containing L(1 ≤ L ≤ 2,000,000,000).

The last test case is followed by a line containing a zero.

Output

For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the length of Bob's luckiest number. If Bob can't construct his luckiest number, print a zero.

Sample Input

8
11
16
0

Sample Output

Case 1: 1
Case 2: 2
Case 3: 0

 

代码：

 

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
typedef long long ll;
const int MAX_N = 10000;
const int MOD = 9901;

ll fpow(ll a, ll p) {
    ll ans = 1;
    while (p) {
        if (p&1) ans = (ans*a) % MOD;
        a = (a*a) % MOD;
        p >>= 1;
    }
    return ans;
}

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

ll m;
ll p[100], c[100];
ll getFactors(ll x)
{
    m = 0;
    for (int i = 1; prime[i] <= x/prime[i]; i++) {
        if (x%prime[i] == 0) {
            p[m] = prime[i];
            c[m] = 0;
            while (x%prime[i] == 0) {
                c[m]++;
                x /= prime[i];
            }
            m++;
        }
    }
    if (x > 1) {
        p[m] = x;
        c[m] = 1;
        m++;
    }
    return m;
}

int main()
{
    getPrime();
    ll A, B;
    cin >> A >> B;
    getFactors(A);
    ll ans = 1;
    for (int i = 0; i < m; i++) {
        if (p[i] % MOD == 1) {
            ans = (B*c[i]+1)%MOD * ans % MOD;
            continue;
        }
        ll a = (fpow(p[i], B*c[i]+1)-1+MOD) % MOD;
        ll b = fpow(p[i]-1, MOD-2);
        ans = ans * a % MOD * b % MOD;
    }
    cout << ans << endl;
    return 0;
}