第九届河南省程序设计大赛—— F Decimal integer conversion

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分类专栏： ========河南省acm竞赛试题======== ========各大OJ题库======== ____南阳理工题库 ========算法与数据结构========

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六岁的小明在学习不同基数之间的数字转换，但他总会在转换过程中写错一个数字。给定他在转换10进制数N为2进制和3进制时的错误结果，你需要找出N的正确值。确保N在1亿范围内且有唯一解。

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题目描述:

XiaoMing likes mathematics, and he is just learning how to convert numbers between different bases , but he keeps making errors since he is only 6 years old. Whenever XiaoMing converts a number to a new base and writes down the result, he always writes one of the digits wrong. For example , if he converts the number 14 into binary (i.e., base 2), the correct result should be "1110", but he might instead write down "0110" or "1111". XiaoMing never accidentally adds or deletes digits, so he might write down a number with a leading digit of " 0" if this is the digit she gets wrong. Given XiaoMing 's output when converting a number N into base 2 and base 3, please determine the correct original value of N (in base 10). (N<=10^10) You can assume N is at most 1 billion, and that there is a unique solution for N.

输入描述:

The first line of the input contains one integers T, which is the nember of test cases (1<=T<=8)
Each test case specifies:
* Line 1: The base-2 representation of N , with one digit written incorrectly.
* Line 2: The base-3 representation of N , with one digit written incorrectly.

输出描述:

For each test case generate a single line containing a single integer , the correct value of N

样例输入:

复制

1
1010
212

样例输出:

题意：输入的两个数一个是2进制，一个是3进制，

并且两个数可能中间某一位是错的，请你找出没错之前对应的十进制数

模拟，将所有情况都试一遍

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<vector>
using namespace std;
string s2,s3;
int l2,l3;
int num2(int i,int j){    //这个函数的作用是，将字符串s2的第i位替换成j，之后返回二进制数s2对应的十进制数 
	int num=0;
	for(int k=0;k<l2;k++){
		if(k==i)num=num*2+j;
		else num=num*2+s2[k]-'0';
	}
	return num;
}
int num3(int i,int j){    //同上个函数一样，区别是这个用来处理三进制数 
	int num=0;
	for(int k=0;k<l3;k++){
		if(k==i)num=num*3+j;
		else num=num*3+s3[k]-'0';
	}
	return num;
}
int main()
{
    int t;
    cin>>t;
	while(t--){
		cin>>s2>>s3; 
		l2=s2.size(),l3=s3.size();
		int flag=0;
		for(int i=0;i<l2;i++){
			for(int j=0;j<2;j++){    //s2共有0——L2-1位，每一位都可能是0或者1 
				int n2=num2(i,j);    //得到换过数之后的s2对应的十进制数 
				for(int i1=0;i1<l3;i1++){
					for(int j1=0;j1<3;j1++){    //得到换过数之后s3对应的十进制数 
						int n3=num3(i1,j1);
						if(n2==n3){    //若相等，则输出，跳出循环 
							cout<<n2<<endl;
				            flag=1;
				            continue;
						}
					}
					if(flag)continue;
				}
				if(flag)continue;
			}
			if(flag)continue;
		}
	}
	return 0;
}