首页 > ACM题库 > HDU-杭电 > hdu 3988-harry potter and the hide story-数论-[解题报告]hoj
2015
04-14

hdu 3988-harry potter and the hide story-数论-[解题报告]hoj

Harry Potter and the Hide Story

Problem Description
iSea is tired of writing the story of Harry Potter, so, lucky you, solving the following problem is enough.

 


Input
The 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 <= 1 000 000 000 000 00
3. 1 <= N <= 1 000 000 000 000 000 000

 


Output
For 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
2 2 2 10 10
 


Sample Output
Case 1: 1 Case 2: 2
 


Author
[email protected]
 


Source
 


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lcy
题目大意:

给定n和k, 求 n! 能被 k^i 整除时,i 的最大取值。

解题思路:

将k分解质因素,问题变为,(1×2×3×…×n) 要被 ( p1^(i*a1) × p2^(i*a2) × … × pn^(i*an) ) 整除,即分子中各分母的质因数的幂次要大于等于分母。

所以根据k的各质因素,求出满足各质因数的幂次 分子>=分母 的关系限制i,算出最大的i即可。

这题要用到unsigned long long,比较坑。。

参考代码:

#include <iostream>
#include <cstring>
#include <cmath>
#define INF 9223372036854775807ULL
using 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;
}

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