algorithm

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:heavy_check_mark: kth_root
(Others/kth_root.hpp)

概要

$x^K \leq N$ を満たす最大の整数 $x$ を返す

使い方

制約

問題例

Verified with

Code

#pragma once
#include <bits/stdc++.h>
#include <cassert>
using namespace std;

// kth root
// N < 2^64, K <= 64
uint64_t kth_root(uint64_t N, uint64_t K = 2) {
    assert(K >= 1);
    if (N <= 1 || K == 1) return N;
    if (K >= 64) return 1;
    if (N == uint64_t(-1)) --N;
    
    auto mul = [&](uint64_t x, uint64_t y) -> uint64_t {
        if (x < UINT_MAX && y < UINT_MAX) return x * y;
        if (x == uint64_t(-1) || y == uint64_t(-1)) return uint64_t(-1);
        return (x <= uint64_t(-1) / y ? x * y : uint64_t(-1));
    };
    auto power = [&](uint64_t x, uint64_t k) -> uint64_t {
        if (k == 0) return 1ULL;
        uint64_t res = 1ULL;
        while (k) {
            if (k & 1) res = mul(res, x);
            x = mul(x, x);
            k >>= 1;
        }
        return res;
    };
    
    uint64_t res;
    if (K == 2) res = sqrtl(N) - 1;
    else if (K == 3) res = cbrt(N) - 1;
    else res = pow(N, nextafter(1 / double(K), 0));
    while (power(res + 1, K) <= N) ++res;
    return res;
}
#line 2 "Others/kth_root.hpp"
#include <bits/stdc++.h>
#line 4 "Others/kth_root.hpp"
using namespace std;

// kth root
// N < 2^64, K <= 64
uint64_t kth_root(uint64_t N, uint64_t K = 2) {
    assert(K >= 1);
    if (N <= 1 || K == 1) return N;
    if (K >= 64) return 1;
    if (N == uint64_t(-1)) --N;
    
    auto mul = [&](uint64_t x, uint64_t y) -> uint64_t {
        if (x < UINT_MAX && y < UINT_MAX) return x * y;
        if (x == uint64_t(-1) || y == uint64_t(-1)) return uint64_t(-1);
        return (x <= uint64_t(-1) / y ? x * y : uint64_t(-1));
    };
    auto power = [&](uint64_t x, uint64_t k) -> uint64_t {
        if (k == 0) return 1ULL;
        uint64_t res = 1ULL;
        while (k) {
            if (k & 1) res = mul(res, x);
            x = mul(x, x);
            k >>= 1;
        }
        return res;
    };
    
    uint64_t res;
    if (K == 2) res = sqrtl(N) - 1;
    else if (K == 3) res = cbrt(N) - 1;
    else res = pow(N, nextafter(1 / double(K), 0));
    while (power(res + 1, K) <= N) ++res;
    return res;
}
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