1408191909-hd-decimal system.cpp

decimal system

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3394 Accepted Submission(s): 1309

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

题目大意

      给出一系列数据，如a(b),表示b进制a,要求出这一系列数据的十进制的和。

 

解题思路

       考的就是一个进制转换，将b进制转换为10进制。如110101(2)转换为10进制为

        sum=1*2^5+1*2^4+0*2^3+1*2^2+0*2^1+1*2^0

知识拓展

       将10进制转换为b进制，是不断取余，然后倒着将余数输出。

代码

<span style="font-size:18px;">#include<stdio.h>
#include<math.h>
int zhuan(int a,int b)
{
	int i;
	double sum;
	sum=0;
	for(i=0;;i++)
	{
		if(a==0)
		    break;
		sum+=a%10*pow(b,i);
		a/=10;
	}
	return sum;
}
int main()
{
	int n;
	int a,b,sum;
	int i;
	while(scanf("%d",&n)==1)
	{
		sum=0;
		for(i=0;i<n;i++)
		{
		    scanf("%d%*c%d%*c",&a,&b);
		    sum+=zhuan(a,b);
		}
		printf("%d\n",sum);
	}
	return 0;
}</span>