http://acm.hdu.edu.cn/showproblem.php?pid=1296

随手练习

 #include <bits/stdc++.h>
using namespace std;
#define ll long long
int n;
char s[];
ll Solve(string s,int i)
{
ll num=;
int l=i-;
int r=i+;
string s1="";
while(s[l]>=''&&s[l]<='')
{
s1+=s[l];
l--;
}
reverse(s1.begin(),s1.end());
ll ans=;
ll sum=;
int len=s1.length();
for(int j=; j<len; j++)
{
ans*=;
ans+=(s1[j]-'');
}
if(ans==){
ans=; }
if(s[r]=='^')
{
r++;
while(s[r]>=''&&s[r]<='')
{
sum*=;
sum+=(s[r]-'');
r++;
}
ll pos=n;
for(int j=; j<sum; j++)
{
pos*=n;
}
num+=pos*ans;
}
else
{
num+=ans*n;
}
i=r;
if(s[l]=='-')
{
num*=-;
}
return num;
}
int main()
{
while(~scanf("%d",&n))
{
ll x=;
scanf("%s",s);
int len=strlen(s);
if((s[]>=''&&s[]<='')||s[]=='X')
{
string s1="";
int i=;
int pos=;
int flag=;
for(i=; s[i]!='+'&&s[i]!='-'&&i<len; i++)
{
s1+=s[i];
if(s[i]=='X')
{
flag=;
pos=i;
}
}
if(flag)
{
x+=Solve(s1,pos);
flag=;
}
else
{
ll ans=;
for(int j=; j<s1.length(); j++)
{
ans*=;
ans+=(s1[j]-'');
}
x+=ans;
}
for(int j=i; j<len;)
{
if(s[j]=='-'||s[j]=='+')
{
string s2="";
s2+=s[j];
j++;
int ans=;
while(s[j]!='-'&&s[j]!='+'&&j<len)
{
s2+=s[j];
ans++;
if(s[j]=='X')
{
pos=ans;
flag=;
}
//cout<<s[j]<<endl;
j++;
}
if(flag)
{
x+=Solve(s2,pos);
flag=;
}
else
{
ll ans=;
for(int j=; j<s2.length(); j++)
{
ans*=;
ans+=(s2[j]-'');
}
if(s2[]=='-')
{
ans*=-;
}
x+=ans;
}
flag=;
}
}
cout<<x<<endl;
}
else
{
int flag=;
int pos=;
string s3="";
s3+=s[];
int i;
for(i=; s[i]!='-'&&s[i]!='+'&&i<len; i++)
{
s3+=s[i];
if(s[i]=='X')
{
flag=;
pos=i;
}
}
if(flag)
{
x+=Solve(s3,pos);
flag=;
}
else
{
ll ans=;
for(int j=; j<s3.length(); j++)
{
ans*=;
ans+=(s3[j]-'');
}
if(s3[]=='-')
{
ans*=-;
}
x+=ans;
}
flag=;
for(int j=i; j<len;)
{
if(s[j]=='-'||s[j]=='+')
{
string s2="";
s2+=s[j];
j++;
int ans=;
while(s[j]!='-'&&s[j]!='+'&&j<len)
{
s2+=s[j];
ans++;
if(s[j]=='X')
{
pos=ans;
flag=;
}
//cout<<s[j]<<endl;
j++;
}
if(flag)
{
x+=Solve(s2,pos);
flag=;
}
else
{
ll ans=;
for(int j=; j<s2.length(); j++)
{
ans*=;
ans+=(s2[j]-'');
}
if(s2[]=='-')
{
ans*=-;
}
x+=ans;
}
flag=;
}
}
cout<<x<<endl;
}
}
return ;
}

HDU1296 Polynomial Problem的更多相关文章

  1. Polynomial Problem(hdu 1296 表达式求值)

    We have learned how to obtain the value of a polynomial when we were a middle school student. If f(x ...

  2. hdu 1296 Polynomial Problem(多项式模拟)

    Problem Description We have learned how to obtain the value of a polynomial when we were a middle sc ...

  3. FZU 2215 Simple Polynomial Problem(简单多项式问题)

    Description 题目描述 You are given an polynomial of x consisting of only addition marks, multiplication ...

  4. FZU2215 Simple Polynomial Problem(中缀表达求值)

    比赛时没做出这题太可惜了. 赛后才反应过来这就是个中缀表达式求值,数字栈存的不是数字而是多项式. 而且,中缀表达式求值很水的,几行就可以搞定. #include<cstdio> #incl ...

  5. 第六届福建省大学生程序设计竞赛(FZU2213—FZU2221)

    from:piaocoder Common Tangents(两圆之间的公公切线) 题目链接: http://acm.fzu.edu.cn/problem.php?pid=2213 解题思路: 告诉你 ...

  6. NOIp初赛题目整理

    NOIp初赛题目整理 这个 blog 用来整理扶苏准备第一轮 csp 时所做的与 csp 没 有 关 系 的历年 noip-J/S 初赛题目,记录了一些我从不知道的细碎知识点,还有一些憨憨题目,不定期 ...

  7. CF #296 (Div. 1) B. Clique Problem 贪心(构造)

    B. Clique Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...

  8. [codeforces 528]B. Clique Problem

    [codeforces 528]B. Clique Problem 试题描述 The clique problem is one of the most well-known NP-complete ...

  9. 周赛-Integration of Polynomial 分类: 比赛 2015-08-02 08:40 10人阅读 评论(0) 收藏

    Integration of Polynomial Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/O ...

随机推荐

  1. js面向(基于)对象编程-三大特征

    ①抽象 js提供以下几种控制方法和属性的访问权限: (1)公开级别:对外公开 (2)私有级别:类本身可以访问,不对外公开 案例如下所示: function Person(name,age,sal){ ...

  2. sql中使用timestamp增量抽取数据

    网址:http://www.cnblogs.com/shuaifei/p/4469526.html 最近的项目中需要对上百万级的数据进行增量抽取操作,因此了解了一下TIMESTAMP的应用,特此记录 ...

  3. 2018.3.3 How too much fructose may cause liver damage

    Fructose is the sweetest of the natural sugars. As its name suggests, it is found mainly in fruits. ...

  4. oracle 11g 常用命令

    sqlplus system/123@ORCL; 查看oracle字符集: select * from nls_database_parameters where parameter ='NLS_CH ...

  5. ACM学习历程——HDU5202 Rikka with string(dfs,回文字符串)

    Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he ...

  6. 【C++ Primer 5th】Chapter 15

    摘要: 1. 面向对象程序设计的核心思想是数据抽象.继承和动态绑定.数据抽象将类的接口和实现分离:继承定义相似的类型并对齐相似关系建模:动态绑定,在一定程度上忽略相似类型的区别,而以统一的方式使用它们 ...

  7. RabbitMQ的持久化机制

    一.问题的引出 RabbitMQ的一大特色是消息的可靠性,那么它是如何保证消息可靠性的呢?——消息持久化.为了保证RabbitMQ在退出,服务重启或者crash等异常情况下,也不会丢失消息,我们可以将 ...

  8. Behave + Selenium(Python)一:

    Behave 介绍:(来自T先生) 最近一个项目用了behave来做测试,因为之前没有接触过,所以写下最近的心得总结. 做自动化的人估计对selenium已经不是很陌生了,但是对于Behave工具,估 ...

  9. MFC中界面自适应

    void CMyDlg::OnSize(UINT nType, int cx, int cy){ CDialogEx::OnSize(nType, cx, cy); CRect rt; GetClie ...

  10. 03_通过OpenHelper获取SqliteDatabase对象

    MyOpenHelper openHelper = new MyOpenHelper(this); 类似于java的File file = new File();只是声明这个东西,但是文件还并没有真正 ...