传送门:http://noi.openjudge.cn/ch0406/1455/

 /-24作业
 //#include "stdafx.h"
 #include<bits/stdc++.h>
 using namespace std;
 int main()
 {
     int n;
     while (cin >> n&&n)
     {
         ] = {  }, s[] = {  }, q = ;
         {//make;
             ), n /= ;
             while (!s[q]) q--;
             ; i <= q; i++) f[i] = s[q - i + ];
             ; i--) ])
             {
                 f[i - ] = , f[i] = ;
                 ;
                 for (int j = i; j <= q; j++)
                     ; }
                 ;
                 break;
             }
         }
         {//back
             ;
             ; i <= q; i++) ans += f[q - i] * pow(, i);
             cout << ans << endl;
         }
     }
     ;
 }

1455:An Easy Problem的更多相关文章

  1. NOI4.6 1455:An Easy Problem

    描述 As we known, data stored in the computers is in binary form. The problem we discuss now is about ...

  2. [openjudge] 1455:An Easy Problem 贪心

    描述As we known, data stored in the computers is in binary form. The problem we discuss now is about t ...

  3. UVA-11991 Easy Problem from Rujia Liu?

    Problem E Easy Problem from Rujia Liu? Though Rujia Liu usually sets hard problems for contests (for ...

  4. An easy problem

    An easy problem Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Sub ...

  5. UVa 11991:Easy Problem from Rujia Liu?(STL练习,map+vector)

    Easy Problem from Rujia Liu? Though Rujia Liu usually sets hard problems for contests (for example, ...

  6. POJ 2826 An Easy Problem?!

    An Easy Problem?! Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7837   Accepted: 1145 ...

  7. hdu 5475 An easy problem(暴力 || 线段树区间单点更新)

    http://acm.hdu.edu.cn/showproblem.php?pid=5475 An easy problem Time Limit: 8000/5000 MS (Java/Others ...

  8. 【暑假】[实用数据结构]UVa11991 Easy Problem from Rujia Liu?

    UVa11991 Easy Problem from Rujia Liu?  思路:  构造数组data,使满足data[v][k]为第k个v的下标.因为不是每一个整数都会出现因此用到map,又因为每 ...

  9. HDU 5475 An easy problem 线段树

    An easy problem Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pi ...

随机推荐

  1. keepalived实现nginx高可用

    keepalived是什么 keepalived直译就是保持存活,在网络里面就是保持在线了,也就是所谓的高可用或热备,用来防止单点故障(单点故障是指一旦某一点出现故障就会导致整个系统架构的不可用)的发 ...

  2. 《剑指Offer》面试题5-替换空格

    题目:请实现一个函数,把字符串中的每个空格替换成"%20".例如输入"We are happy.",则输出"We%20are%20happy.&quo ...

  3. web前端开发初学者必看的学习路线(附思维导图)

    很多同学想学习WEB前端开发,虽然互联网有很多的教程.网站.书籍,可是却又不知从何开始如何选取.看完网友高等游民白乌鸦无私分享的原标题为<写给同事的前端学习路线>这篇文章,相信你会有所收获 ...

  4. linux 下gcc生成intel汇编

    留作备忘: gcc -S -masm=intel xxxx.c 生成elf可执行文件: gcc -o xxx xxxx.s 反汇编 objdump xxx 补充: 在使用gcc 对C语言程序进行编译时 ...

  5. Hadoop(十四)MapReduce原理分析

    前言 上一篇我们分析了一个MapReduce在执行中的一些细节问题,这一篇分享的是MapReduce并行处理的基本过程和原理. Mapreduce是一个分布式运算程序的编程框架,是用户开发“基于had ...

  6. 242. Valid Anagram(leetcode)

    Given two strings s and t, write a function to determine if t is an anagram of s. For example, s = & ...

  7. LeetCode 229. Majority Element II (众数之二)

    Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorit ...

  8. LeetCode 33. Search in Rotated Sorted Array(在旋转有序序列中搜索)

    Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e. ...

  9. struts2(一)之初识struts2

    前言 我们都知道struts2是一个框架,那什么是框架呢?很多人其实不太明白,其实框架就是一个半成品,别人将一些功能已经写好了,我们只需要拿来用即可,像我们之前 使用的dbutils框架,操作数据,只 ...

  10. sqoop1.9.7安装和使用

    安装1.下载sqoop1.9.7.地址: http://www.apache.org/dyn/closer.lua/sqoop/1.99.72.解压sqoop ,并配置环境变量 ~/.bash_pro ...