POJ 刷题系列:2109. Power of Cryptography
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POJ 刷题系列:2109. Power of Cryptography
传送门:2109. Power of Cryptography
题意:
给定n,p,求k,使得
kn=p
思路:
这不应该放在贪心里啊!!!刷新了我对double的认识,实际上double的表示范围是巨大的。有
浮点型: Float 32位单精度浮点数 10^-38~10^38和-10^-38~-10^38Double 64位双精度浮点数 10^-308~10^308和-10^-308~-10^308
代码如下:
import java.io.BufferedReader;import java.io.File;import java.io.FileInputStream;import java.io.IOException;import java.io.InputStream;import java.io.InputStreamReader;import java.io.PrintWriter;import java.util.Arrays;import java.util.Map;import java.util.StringTokenizer;public class Main{ String INPUT = "./data/judge/201712/P2109.txt"; public static void main(String[] args) throws IOException { new Main().run(); } void read() { while (more()) { double n = nd(); double p = nd(); double k = Math.pow(p, 1 / n); out.printf("%.0f\n", k); } } FastScanner in; PrintWriter out; void run() throws IOException { boolean oj; try { oj = ! System.getProperty("user.dir").equals("F:\\oxygen_workspace\\Algorithm"); } catch (Exception e) { oj = System.getProperty("ONLINE_JUDGE") != null; } InputStream is = oj ? System.in : new FileInputStream(new File(INPUT)); in = new FastScanner(is); out = new PrintWriter(System.out); long s = System.currentTimeMillis(); read(); out.flush(); if (!oj){ System.out.println("[" + (System.currentTimeMillis() - s) + "ms]"); } } public boolean more(){ return in.hasNext(); } public int ni(){ return in.nextInt(); } public long nl(){ return in.nextLong(); } public double nd(){ return in.nextDouble(); } public String ns(){ return in.nextString(); } public char nc(){ return in.nextChar(); } class FastScanner { BufferedReader br; StringTokenizer st; boolean hasNext; public FastScanner(InputStream is) throws IOException { br = new BufferedReader(new InputStreamReader(is)); hasNext = true; } public String nextToken() { while (st == null || !st.hasMoreTokens()) { try { st = new StringTokenizer(br.readLine()); } catch (Exception e) { hasNext = false; return "##"; } } return st.nextToken(); } String next = null; public boolean hasNext(){ next = nextToken(); return hasNext; } public int nextInt() { if (next == null){ hasNext(); } String more = next; next = null; return Integer.parseInt(more); } public long nextLong() { if (next == null){ hasNext(); } String more = next; next = null; return Long.parseLong(more); } public double nextDouble() { if (next == null){ hasNext(); } String more = next; next = null; return Double.parseDouble(more); } public String nextString(){ if (next == null){ hasNext(); } String more = next; next = null; return more; } public char nextChar(){ if (next == null){ hasNext(); } String more = next; next = null; return more.charAt(0); } } static class D{ public static void pp(int[][] board, int row, int col) { StringBuilder sb = new StringBuilder(); for (int i = 0; i < row; ++i) { for (int j = 0; j < col; ++j) { sb.append(board[i][j] + (j + 1 == col ? "\n" : " ")); } } System.out.println(sb.toString()); } public static void pp(char[][] board, int row, int col) { StringBuilder sb = new StringBuilder(); for (int i = 0; i < row; ++i) { for (int j = 0; j < col; ++j) { sb.append(board[i][j] + (j + 1 == col ? "\n" : " ")); } } System.out.println(sb.toString()); } } static class ArrayUtils { public static void fill(int[][] f, int value) { for (int i = 0; i < f.length; ++i) { Arrays.fill(f[i], value); } } public static void fill(int[][][] f, int value) { for (int i = 0; i < f.length; ++i) { fill(f[i], value); } } public static void fill(int[][][][] f, int value) { for (int i = 0; i < f.length; ++i) { fill(f[i], value); } } } static class Num{ public static <K> void inc(Map<K, Integer> mem, K k) { if (!mem.containsKey(k)) mem.put(k, 0); mem.put(k, mem.get(k) + 1); } }}
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