decimal system

浅浅的花香味﹌ 2023-07-17 04:57 25阅读 0赞

## 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 <>,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

**Code:**  
思路：利用getline()函数一次读入一行字符串，将该字符串的内容分为x和y，然后将y进制的数x转化为对应的十进制数（乘基相加）进行求和计算

#include<iostream>
    #include<string>
    #include<math.h>
    using namespace std;
    //将y进制表示的数x转化为十进制数
    int convert(int x, int y) {
        
     	int data, deci = 0, i = 0;
     	while (x != 0) {
        
      		data = x % 10;
      		deci = deci + data * pow(y, i);
      		i++;
      		x = x / 10;
    	 }
     	return deci;
     //deci的值即为最终转化成的十进制数
    }
    int main() {
        
    	int n;	
    	int sum;
    	string s, s1, s2, str1;		
    	while (cin >> n) {
        
     		if (n <= 0 || n > 100) return 0;
     		sum = 0;
     		getline(cin, str1);//消除cin的影响
     		for (int i = 0; i < n; i++) {
        
      			getline(cin, s);// 需要在头文件中加<string>,一次读入一行字符串，可读入空格
      			s1 = s.substr(0, s.find('('));//s.substr（pos,count）函数,用户返回字符串s中从第pos个字符开始，
                //长为count的子串;find()函数用于返回字符在母串中的位置
     			s2 = s.substr(s.find('(') + 1, s.find(')') - s.find(')') - 1);
     			int x = atoi(s1.c_str());//s1.c_str()返回一个指向字符串的指针；
             //atoi(s1.c_str())把字符串转化为整型
     			int y = atoi(s2.c_str());
     			if (y <= 1 || y > 10) return 0;
     			sum = sum + convert(x, y);
      		}
     		if (sum > 100000000) return 0;
      		cout << sum << endl;	
    	}
    }