1010. Radix (25)(出错较多待改进)
Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is "yes", if 6 is a decimal number and 110 is a binary number.
Now for any pair of positive integers N1 and N2, your task is to find the radix of one number while that of the other is given.
Input Specification:
Each input file contains one test case. Each case occupies a line which contains 4 positive integers:
N1 N2 tag radix
Here N1 and N2 each has no more than 10 digits. A digit is less than its radix and is chosen from the set {0-9, a-z} where 0-9 represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. The last number "radix" is the radix of N1 if "tag" is 1, or of N2 if "tag" is 2.
Output Specification:
For each test case, print in one line the radix of the other number so that the equation N1 = N2 is true. If the equation is impossible, print "Impossible". If the solution is not unique, output the smallest possible radix.
Sample Input 1:
6 110 1 10
Sample Output 1:
2
Sample Input 2:
1 ab 1 2
Sample Output 2:
Impossible
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
long long binarySearch();
int cmp( long long k);
char a[],b[],c[];
long long ans[];
long long low,high,len,valuea;
int main()
{
int tag;
long long radixa,temp,ret;
int i;
scanf("%s %s %d %lld",a,b,&tag,&radixa);
if( tag==)
{
strcpy(c,a);
strcpy(a,b);
strcpy(b,a);
}
for( i=; a[i]!='\0'; i++)
{
if( a[i]>='' && a[i]<='')
temp = a[i]-'';
else if( a[i]>'a' && a[i]<'z')
temp = a[i]-'a'+;
valuea = valuea*radixa + temp;
} for( i=; b[i]!='\0'; i++)
{
if( b[i]>='' && b[i]<='')
temp = b[i]-'';
else if( b[i]>'a' && b[i]<'z')
temp = b[i]-'a'+;
ans[i]=temp;
if( low<temp)
low = temp;
}
low++;
len = strlen(b);
if( low>valuea)
high = low+;
else high=valuea+;
ret = binarySearch();
if( ret==-)
printf("Impossible\n");
else printf("%lld\n",ret);
return ;
}
long long binarySearch()
{
long long l=low,h=high,mid;
while( l<=h )
{
mid = (l+h)/;
if(cmp(mid)==)
return mid;
else if(cmp(mid)<)
l= mid+;
else h=mid-;
}
return -;
} int cmp( long long k)
{
long long valueb=;
int i;
for( i=; i<len; i++)
valueb = k*valueb+ans[i];
if( valueb< || valueb>valuea)
return ;
else if( valueb<valuea)
return -;
else if ( valuea==valueb)
return ;
}
1010. Radix (25)(出错较多待改进)的更多相关文章
- PAT 解题报告 1010. Radix (25)
1010. Radix (25) Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 11 ...
- PAT 甲级 1010 Radix (25)(25 分)进制匹配(听说要用二分,历经坎坷,终于AC)
1010 Radix (25)(25 分) Given a pair of positive integers, for example, 6 and 110, can this equation 6 ...
- pat 甲级 1010. Radix (25)
1010. Radix (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Given a pair of ...
- 已经菜到不行了 PAT 1010. Radix (25)
https://www.patest.cn/contests/pat-a-practise/1010 题目大意: 输入四个数字,a,b,c,d. a和b是两个数字,c=1表示是第一个数字,c=2表示是 ...
- 1010. Radix (25)(未完成)
Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The an ...
- 1010. Radix (25)
Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The an ...
- PAT (Advanced Level) 1010. Radix (25)
撸完这题,感觉被掏空. 由于进制可能大的飞起..所以需要开longlong存,答案可以二分得到. 进制很大,导致转换成10进制的时候可能爆long long,在二分的时候,如果溢出了,那么上界=mid ...
- 1010. Radix (25) pat
Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The an ...
- 1010 Radix (25)(25 point(s))
problem Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true ...
随机推荐
- Ionic之ui-sref引入图片,图片部分挡住解决方案
ionic图片设置大小跟图片像素相同,但是使用ui-sref="parkInfo"上半部分图片会挡住,增加height的高度,就可以显示原本图片 页面: <ion-conte ...
- Java关键字-volatile
关键字volatile可以说是Java虚拟机提供的最轻量级的同步机制. 一旦某个共享变量(类的成员变量.类的静态成员变量)被volatile修饰之后,那么就具备了两层语义: 1.保证了不同线程对这个变 ...
- 学JAVA第二十四天,Set集合与StringBuilder
下面的内容需要慢慢看,因为,我的语言表达能力不是很好 首先说Set把,Set集合是一个无序且不允许重复的集合,而且查找效率也是快的可怕的. 但是,有些时候,我们必须要用储存多个相同的值时,Set也是可 ...
- Java Web开发之Spring | SpringMvc | Mybatis | Hibernate整合、配置、使用
1.Spring与Mybatis整合 web.xml: <?xml version="1.0" encoding="UTF-8"?> <web ...
- RHEL5.8上SAMBA源码修改打包安装流程
之前一直使用系统自带的SAMBA,近期需要对SAMBA代码做一些修改,然后还是打算用RPM包的方式来安装部署. 这个流程本身不复杂,在这里记录下来,免得在另外写说明文档. 关键词:RHEL5.8, s ...
- JDBC ResultSet分析
JDBC1.0 .JDBC2.0 .JDBC3.0 中分别用以下方法创建Statement . JDBC1.0 : createStatement() JDBC2.0 : createStatemen ...
- (转)Spring如何装配各种集合类型的属性
http://blog.csdn.net/yerenyuan_pku/article/details/52858499 在前面我们已经会注入基本类型对象和其他bean,现在我们就来学习如何注入各种集合 ...
- [整理]ADB命令行学习笔记
global driver# 元素定位driver.find_element_by_id("id") # id定位driver.find_element_by_name(" ...
- 7-Java-C(搭积木)
题目描述: 小明最近喜欢搭数字积木, 一共有10块积木,每个积木上有一个数字,0~9. 搭积木规则: 每个积木放到其它两个积木的上面,并且一定比下面的两个积木数字小. 最后搭成4层的金字塔形,必须用完 ...
- uva1228 Integer Transmission
这道题思维很灵活.也有点套路的意思. 首先规定0,1分别按照原来的顺序接收,只是01换位.这样简化了思维.(否则并不会有更优结果它.,比较好想)最大值和最小值可以贪心得到.那么接下来就是给定一个整数P ...