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1 // 16_10jinzhi.cpp : 定义控制台应用程序的入口点。 2 // 3 4 #include "stdafx.h" 5 #include<iostream> 6 #include<string> 7 #include<math.h> 8 using namespace std; 9 10 11 int getNum(char c) 12 { 13 int temp = 0; 14 switch(c) 15 { 16 case ‘A‘: 17 temp = 10; 18 break; 19 case ‘B‘: 20 temp = 11; 21 break; 22 case ‘C‘: 23 temp = 12; 24 break; 25 case ‘D‘: 26 temp = 13; 27 break; 28 case ‘E‘: 29 temp = 14; 30 break; 31 case ‘F‘: 32 temp = 15; 33 break; 34 case ‘0‘: 35 temp = 0; 36 break; 37 case ‘1‘: 38 temp = 1; 39 break; 40 case ‘2‘: 41 temp = 2; 42 break; 43 case ‘3‘: 44 temp = 3; 45 break; 46 case ‘4‘: 47 temp = 4; 48 break; 49 case ‘5‘: 50 temp = 5; 51 break; 52 case ‘6‘: 53 temp = 6; 54 break; 55 case ‘7‘: 56 temp = 7; 57 break; 58 case ‘8‘: 59 temp = 8; 60 break; 61 case ‘9‘: 62 temp = 9; 63 break; 64 } 65 return temp; 66 } 67 68 int _tmain(int argc, _TCHAR* argv[]) 69 { 70 string OX_num; 71 while(cin>>OX_num) 72 { 73 int len = OX_num.length(); 74 if(len <= 2) 75 return -1; 76 int result = 0; 77 for(int i = 2; i < len;i++) 78 { 79 int temp = 0; 80 if((OX_num[i]>=‘A‘ && OX_num[i] <= ‘F‘)|| (OX_num[i]>=‘0‘ && OX_num[i] <= ‘9‘) ) 81 temp = getNum(OX_num[i]); 82 else 83 return -1; 84 85 result += temp*(pow(static_cast<double>(16),len - i - 1)); 86 } 87 cout<<result<<endl; 88 } 89 return 0; 90 }
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原文地址:http://www.cnblogs.com/lp3318/p/5840515.html