题目链接:http://acm.pku.edu.cn/JudgeOnline/problem?id=1555

#include <stdio.h>

int coe[9];
int main(int argc, char* argv[])
{
 while(scanf("%d",&coe[0]) != EOF)
 {
  int i;
  for(i = 1;i < 9;i++)
   scanf("%d",&coe[i]);

bool first = true;;
  for(i = 0;i < 9;i++)
  {
   if(first)
   {
    if(coe[i] == 0)
     continue;
    else
    {
     if(8 - i > 1)
     {
      if(coe[i] == 1)
      {
       printf("%s%d","x^",8 - i);
      }
      else if(coe[i] == -1)
      {
       printf("%s%d","-x^",8 - i);
      }
      else if(coe[i] > 1)
      {
       printf("%d%s%d",coe[i],"x^",8 - i);
      }
      else if(coe[i] < -1)
      {
       printf("%d%s%d",coe[i],"x^",8 - i);
      }
     }
     else if(8 - i == 1)
     {
      if(coe[i] == 1)
      {
       printf("%s","x");
      }
      else if(coe[i] == -1)
      {
       printf("%s","-x");
      }
      else if(coe[i] > 1)
      {
       printf("%d%s",coe[i],"x");
      }
      else if(coe[i] < -1)
      {
       printf("%d%s",coe[i],"x");
      }
     }
     else if(8 - i == 0)
     {
      printf("%d",coe[i]);
     }
     first = false;
    }
   }
   else
   {
    if(8 - i > 1)
    {
     if(coe[i] == 1)
     {
      printf("%s%d"," + x^",8 - i);
     }
     else if(coe[i] == -1)
     {
      printf("%s%d"," - x^",8 - i);
     }
     else if(coe[i] > 1)
     {
      printf("%s%d%s%d"," + ",coe[i],"x^",8 - i);
     }
     else if(coe[i] < -1)
     {
      printf("%s%d%s%d"," - ",-coe[i],"x^",8 - i);
     }
    }
    else if(8 - i == 1)
    {
     if(coe[i] == 1)
     {
      printf("%s"," + x");
     }
     else if(coe[i] == -1)
     {
      printf("%s"," - x");
     }
     else if(coe[i] > 1)
     {
      printf("%s%d%s"," + ",coe[i],"x");
     }
     else if(coe[i] < -1)
     {
      printf("%s%d%s"," - ",-coe[i],"x");
     }
    }
    else if(8 - i == 0)
    {
     if(coe[i] > 0)
     {
      printf(" + %d",coe[i]);
     }
     else if(coe[i] < 0)
     {
      printf(" - %d",-coe[i]);
     }
    }
   }
  }
  if(first)
   printf("0");
  printf("/n");
 }
 return 0;
}

BD_source code for problem 1555相关推荐

  1. Code Jam Problem D. GoroSort

    题目来源:https://code.google.com/codejam/contest/dashboard?c=975485#s=p3 题目感觉非常不错,有思考的深度! 可以参考官方给出的解析:ht ...

  2. 安全研究 # Neural Network-based Graph Embedding for Cross-Platform Binary Code Similarity Detection

    论文分享<Neural Network-based Graph Embedding for Cross-Platform Binary Code Similarity Detection> ...

  3. js promises 踩坑 填坑 We have a problem with promises

    We have a problem with promises promise 填坑 对于 promise return 与否,结果真的不一样哦. By: Nolan Lawson Published ...

  4. BZOJ1555 KD之死

    如果没有必选的限制条件,就是水题了... 只要按照w + t排序就可以了,然后搞个堆来维护 于是有了限制条件,还是水题... 到了必选的时候强制选上,不加入堆中即可. 1 /************* ...

  5. 对Oracle中索引叶块分裂而引起延迟情况的测试和分析

    在版本10.2.0.4未打上相关one-off补丁的情况下,分别对ASSM和MSSM管理模式表空间进行索引分裂测试,经过测试的结论如下: l  在10gr2版本中MSSM方式是不能避免索引分裂引起交易 ...

  6. how-to-get-a-job-in-deep-learning

    http://blog.deepgram.com/how-to-get-a-job-in-deep-learning/ How to Get a Job In Deep Learning 22 SEP ...

  7. 2018 东北地区大学生程序设计竞赛(ABEHIK)

    HDU6500:Problem A. Game with string 题意: 给你一个字符串s以及它的m个子串的首尾位置,现在Alice和 Bob两个人轮流在任一子串的前面或者后面加1个字符,要求加 ...

  8. 【POJ】【2449】Remmarguts' Date

    K短路/A* 经(luo)典(ti) K短路题目= = K短路学习:http://www.cnblogs.com/Hilda/p/3226692.html 流程: 先把所有边逆向,做一遍dijkstr ...

  9. NDC 2010视频下载:看看其他微软平台程序员们都在做什么

    原文地址:<NDC 2010视频下载:看看其他微软平台程序员们都在做什么> NDC(Norwegian Developers Conference,挪威开发者大会)是一年一度的挪威最大的微 ...

  10. COMP0037 Coursework Investigating Path Planning Algorithms

    COMP0037 Coursework 1 Term 2, 2019 "Path Planning in a Known World" Investigating Path Pla ...

最新文章

  1. PHP6 Web 开发读书笔记
  2. SQLSERVER执行计划详解
  3. CSS中list-style详解
  4. java定时执行一次_java Timer(定时调用、实现固定时间执行)
  5. Java语言学习概述
  6. java 方法 示例_Java扫描器具有示例的NextNextShort()方法
  7. 打印更无缝:微软改善Win11中通用打印体验
  8. 【IT旁观者】喜羊羊与唐三藏
  9. python一个函数调用另一个函数的返回值_python-调用另一个函数后立即从函数返回...
  10. MIT联手IBM发布超大数据集:100多万短视频,多维度标注
  11. 内存释放_把微信这两个开关关掉, 手机立马释放大量内存, 快去试试吧
  12. docker学习笔记3:镜像操作(查找和下载)
  13. 基于Tesseract的OCR识别--身份证
  14. windows上面运行jar文件,Windows服务器后台运行jar包
  15. DELL EMC Isilon配额Quota
  16. java 爬取网页的数据_java爬取网页数据
  17. STM32超低功耗入门之低功耗运行模式
  18. Automatic Panoramic Image Stitching using Invariant Features 【笔记】
  19. 查询活动开始时间和结束时间
  20. python统计小说人物_python统计喜欢的小说主角出场次数

热门文章

  1. lighttpd + php for android
  2. 5.1--5.1.2 Permutations of a Multiset 多重集合的排列
  3. css 列表内容滚动,stroll.js 炫酷的 CSS3 列表滚动动画集合 - 文章教程
  4. 今天我理解了妈妈为什么会被电信诈骗
  5. Confluence 6 为站点启用匿名用户访问
  6. GoldenGate 1403错误解决方法
  7. HashTable 简述
  8. CO02工单下达时错误“订单类型 ZP91 工厂 1000: 没有检查工序的维护规则”
  9. AVFoundation总结
  10. 彻底解决四种路由器不能上网故障