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  • 18.12.21 【USACO】Times17

    题目描述

    After realizing that there is much money to be made in software development, Farmer John has launched a small side business writing short
    programs for clients in the local farming industry. 

    Farmer John's first programming task seems quite simple to him -- almost too simple: his client wants him to write a program that takes a number N as input, and prints 17 times N as output.  Farmer John has just finished writing this simple program when the client calls him up in a panic and informs him that the input and output both must be expressed as binary numbers, and that these might be quite large.

    Please help Farmer John complete his programming task.  Given an input number N, written in binary with at most 1000 digits, please write out the binary representation of 17 times N.

    输入

    * Line 1: The binary representation of N (at most 1000 digits).

    输出

    * Line 1: The binary representation of N times 17.

    样例输入

    10110111

    样例输出

    110000100111

    来源

    USACO 2012 March Bronze

     1 #include <iostream>
     2 #include <string.h>
     3 #include <algorithm>
     4 #include <stack>
     5 #include <string>
     6 #include <math.h>
     7 #include <queue>
     8 #include <stdio.h>
     9 #include <string.h>
    10 #include <set>
    11 #include <vector>
    12 #include <fstream>
    13 #define maxn 1005
    14 #define inf 999999
    15 #define cha 127
    16 #define eps 1e-6 
    17 #define oo 1503
    18 using namespace std;
    19 
    20 string a, b;
    21 int n1[maxn], n2[maxn];
    22 
    23 void init() {
    24     cin >> a;
    25     b = a + "0000";
    26     int sizea = a.length(), sizeb = b.length();
    27     for (int i = 4; i < sizeb; i++)
    28         n1[i] = a[i-4] - '0';
    29     for (int i = 0; i < sizeb; i++)
    30         n2[i] = b[i] - '0';
    31     for (int i = sizeb - 1; i >= 1; i--) {
    32         n2[i] = n2[i] + n1[i];
    33         n2[i - 1] += n2[i] / 2;
    34         n2[i] %= 2;
    35     }
    36     string tmp;
    37     while (n2[0] != 0) {
    38         char ch = n2[0] % 2 + '0';
    39         tmp = ch + tmp;
    40         n2[0] /= 2;
    41     }
    42     cout << tmp;
    43     for (int i = 1; i < sizeb; i++)
    44         printf("%d", n2[i]);
    45     printf("
    ");
    46 }
    47 
    48 int main() {
    49     init();
    50     return 0;
    51 }
    View Code
    注定失败的战争,也要拼尽全力去打赢它; 就算输,也要输得足够漂亮。
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  • 原文地址:https://www.cnblogs.com/yalphait/p/10155117.html
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