AtCoder Beginner Contest 180 Participation Report

AtCoder Beginner Contest 180 Participation Report

ABC180A - box

Break through in 1 minute. Just write.

N, A, B = map(int, input().split())

print(N - A + B)

ABC180B - Various distances

Break through in 3 minutes. Just write.

N, *x = map(int, open(0).read().split())

print(sum(abs(a) for a in x))
print(sum(a * a for a in x) ** 0.5)
print(max(abs(a) for a in x))

ABC180C - Cream puff

Break through in 3 minutes. Just write. If you know that you can turn to the square root of N, it shouldn't be difficult.

N = int(input())

result = set()
for i in range(1, int(N ** 0.5) + 1):
    if N % i == 0:
        result.add(i)
        result.add(N // i)
print(*sorted(result), sep='\n')

ABC180D - Takahashi Unevolved

Break through in 14 minutes. It takes too long. It is clearer than looking at the fire that a naive code that simulates obediently will result in a TLE when B is small. When it comes to * A is better than + B. Settle on the conclusion that you should just do * A while the increment is small and add the rest together. Since A is at least 2, you can simulate it obediently * O * (log N) </ i>) So there is no problem.

X, Y, A, B = map(int, input().split())

result = 0
while X * A < Y and X * A < X + B:
    X *= A
    result += 1
result += ((Y - 1) - X) // B
print(result)

Postscript: I thought that there were many people who got penalized due to the D problem, but if I write it without devising, it is because X * A overflows even with int64. I see.

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

func main() {
	defer flush()

	X := readInt()
	Y := readInt()
	A := readInt()
	B := readInt()

	result := 0
	for X <= (Y-1)/A && X*A < X+B {
		X *= A
		result++
	}
	result += ((Y - 1) - X) / B
	println(result)
}

const (
	ioBufferSize = 1 * 1024 * 1024 // 1 MB
)

var stdinScanner = func() *bufio.Scanner {
	result := bufio.NewScanner(os.Stdin)
	result.Buffer(make([]byte, ioBufferSize), ioBufferSize)
	result.Split(bufio.ScanWords)
	return result
}()

func readString() string {
	stdinScanner.Scan()
	return stdinScanner.Text()
}

func readInt() int {
	result, err := strconv.Atoi(readString())
	if err != nil {
		panic(err)
	}
	return result
}

var stdoutWriter = bufio.NewWriter(os.Stdout)

func flush() {
	stdoutWriter.Flush()
}

func println(args ...interface{}) (int, error) {
	return fmt.Fprintln(stdoutWriter, args...)
}

ABC180E - Traveling Salesman among Aerial Cities

It broke through in 39 and a half minutes. I solved it by copying the code of the ant book. I wonder if I could get the yellow performance if I copied it, shit.

#include <bits/stdc++.h>
#define rep(i, a) for (int i = (int)0; i < (int)a; ++i)
using ll = long long;
using namespace std;

#define MAX_N 17
#define INF 2147483647

ll N;
ll dp[1 << MAX_N][MAX_N];
ll d[MAX_N][MAX_N];

void solve() {
    rep(i, 1 << N) rep(j, N) dp[i][j] = INF;
    dp[(1 << N) - 1][0] = 0;

    for (int S = (1 << N) - 2; S >= 0; S--) {
        for (int v = 0; v < N; v++) {
            for (int u = 0; u < N; u++) {
                dp[S][v] = min(dp[S][v], dp[S | 1 << u][u] + d[v][u]);
            }
        }
    }
    cout << dp[0][0] << endl;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    cin >> N;

    vector<ll> X(N), Y(N), Z(N);
    rep(i, N) {
        cin >> X[i] >> Y[i] >> Z[i];
    }

    rep(i, N) {
        rep(j, N) {
            d[i][j] = abs(X[i] - X[j]) + abs(Y[i] - Y[j]) + max(0ll, Z[i] - Z[j]);
        }
    }

    solve();

    return 0;
}

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