HDU 3988 Harry Potter and the hide story (number theory-integer and prime number)

Harry Potter and the hide storyproblem descriptionisea is tired of writing the story of Harry Potter, so, lucky you, solving the following problem is enough.

 
Inputthe first line contains a single integer T, indicating the number of test cases.
Each test case contains two integers, N and K.

Technical Specification

1. 1 <= T <= 500
2. 1 <= k <= 1 000 000 000 00
3. 1 <= n <= 1 000 000 000 000 000
 
Outputfor each test case, output the case number first, then the answer, if the answer is bigger than 9 223 372 036 854 775 807, output "INF" (without quote ).
Sample Input

22 210 10

 
Sample output

Case 1: 1Case 2: 2

 
Author [email protected]
Source2011 multi-university training contest 15-host by whu
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Question:

Given N and K, evaluate n! Maximum Value of I when it can be divisible by K ^ I.

Solution:

Change K into a qualitative factor, (1 × 2 × 3 ×... ×n) to be (P1 ^ (I * A1) × P2 ^ (I * A2) ×... ×pn ^ (I * an) Division, that is, the power of the prime factor of each denominator in the numerator is greater than or equal to the denominator.

Therefore, based on each qualitative factor of K, find the power numerator that satisfies each prime factor> = the limit of the denominator relationship I, and calculate the maximum I.

Unsigned long is used for this question ..

Reference code:

#include <iostream>#include <cstring>#include <cmath>#define INF 9223372036854775807ULLusing namespace std;typedef unsigned long long ull;const int MAXN = 10000010;int T, cnt;ull N, K, ans, factorA[MAXN], factorB[MAXN], totFactor, prime[MAXN], totPrime;bool isPrime[MAXN];void getPrime(ull n) {    memset(isPrime, true, sizeof(isPrime));    totPrime = 0;    for (ull i = 2; i <= n; i++) {        if (isPrime[i]) {            prime[++totPrime] = i;        }        for (ull j = 1; j <= totPrime && i*prime[j] <= n; j++) {            isPrime[i*prime[j]] = false;            if (i % prime[j] == 0) break;        }    }}void getFactor(ull n) {    /*    ull now = n;    totFactor = 0;    for (ull i = 2; i*i <= n; i++) {        if (now % i == 0) {            factorA[++totFactor] = i;            factorB[totFactor] = 0;            while (now % i == 0) {                factorB[totFactor]++;                now /= i;            }        }    }    if (now != 1) {        factorA[++totFactor] = now;        factorB[totFactor] = 1;    }    */    totFactor = 0;    ull now = n;    for (ull i = 1; i <= totPrime && prime[i] <= now; i++) {        if (now % prime[i] == 0) {            factorA[++totFactor] = prime[i];            factorB[totFactor] = 0;            while (now % prime[i] == 0) {                factorB[totFactor]++;                now /= prime[i];            }        }    }    if (now != 1) {        factorA[++totFactor] = now;        factorB[totFactor] = 1;    }}void solve() {    if (K == 1) {        cout << "Case " << ++cnt << ": inf" << endl;    } else {        getFactor(K);        ans = INF;        for (ull i = 1; i <= totFactor; i++) {            ull temp = N, sum = 0;            while (temp > 0) {                sum += temp / factorA[i];                temp /= factorA[i];            }            if (sum / factorB[i] < ans) {                ans = sum / factorB[i];            }        }        cout << "Case " << ++cnt << ": " << ans << endl;    }}int main() {    ios::sync_with_stdio(false);    cin >> T;    getPrime(10000000);    while (T--) {        cin >> N >> K;        solve();    }    return 0;}