2106:decimal system

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#numbers #system #output #each #import #input

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本文介绍了一个简单的程序，用于将不同进制的数值转换为十进制，并计算这些数值的总和。该程序能够处理从二进制到十进制之间的各种数制，适用于计算机科学学习者理解数制转换原理。

Problem Description

As we know , we always use the decimal system in our common life, even using the computer. If we want to calculate the value that 3 plus 9, we just import 3 and 9.after calculation of computer, we will get the result of 12.
But after learning <<The Principle Of Computer>>,we know that the computer will do the calculation as the following steps:
1 computer change the 3 into binary formality like 11;
2 computer change the 9 into binary formality like 1001;
3 computer plus the two number and get the result 1100;
4 computer change the result into decimal formality like 12;
5 computer export the result;

In the computer system there are other formalities to deal with the number such as hexadecimal. Now I will give several number with a kind of change method, for example, if I give you 1011(2), it means 1011 is a number in the binary system, and 123(10) means 123 if a number in the decimal system. Now I will give you some numbers with any kind of system, you guys should tell me the sum of the number in the decimal system.

Input

There will be several cases. The first line of each case contains one integers N, and N means there will be N numbers to import, then there will be N numbers at the next N lines, each line contains a number with such form : X1….Xn.(Y), and 0<=Xi<Y, 1<Y<=10. I promise you that the sum will not exceed the 100000000, and there will be at most 100 cases and the 0<N<=1000.

Output

There is only one line output case for each input case, which is the sum of all the number. The sum must be expressed using the decimal system.

Sample Input

3
1(2)
2(3)
3(4)

4
11(10)
11(2)
11(3)
11(4)

Sample Output

6
23

分析：

实际上就是任意进制向十进制转化的问题。

代码如下：

#include <iostream>
using namespace std;

int pow(int m,int n)//求m的n次幂
{
	long int pro = 1;
	while(n--)
		pro *= m;
	return pro;
}

int main()
{
	int T,p,q;//k为指数
	
	char left_arc,right_arc;
	while (cin >> T)
	{	
		int sum = 0;
		while(T--)
		{
			int k = 0;
			cin >> p >> left_arc >> q >> right_arc;

			while (p != 0)
			{
				sum += p % 10 * pow(q,k++);
				p /= 10;
			}		
		}
		cout << sum << endl;
	}

	return 0;
}