ZD_source code for problem 2971

题目链接:http://acm.zju.edu.cn/show_problem.php?pid=2971

解题心得:

1，用scanf("%d",&i)接受整型后，再用gets接受带空格的字符串之前要调用getchar将/n滤掉

2，用关键字是字符串的map时要用string类型

#include <stdio.h>
#include <map>
#include <vector>
#include <string>
using namespace std;

map<string,int> times;
map<string,int> str2int;
vector<string> number;

string str[] = {"zero","one","two","three","four","five","six","seven","eight","nine",
    "ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen",
    "eighteen","nineteen","twenty","thirty","forty","fifty","sixty","seventy",
    "eighty","ninety"};
void Input()
{
 number.clear();
 string temp;
 char input[128];
 gets(input);

 int i;
 for(i = 0;input[i] != '/0';i++)
 {
  if(input[i] != ' ')
  {
   temp += input[i];
  }
  else
  {
   number.push_back(temp);
   temp = "";
  }
 }
 number.push_back(temp);
}

void Init()
{
 int i;
 for(i = 0;i <= 20;i++)
  str2int[str[i]] = i;
 
 int j;
 for(i = 21,j = 30;i < 28;i++,j += 10)
  str2int[str[i]] = j;

 times[string("and")] = 1;
 times[string("hundred")] = 100;
 times[string("thousand")] = 1000;
 times[string("million")] = 1000000;
}

void Work()
{
 int size = number.size();
 int i,temp,result;
 for(i = 0,temp = 0,result = 0;i < size;i++)
 {
  if(times.find(number[i]) != times.end())
  {
   temp *= times[number[i]];
   if(number[i] == "million" || number[i] == "thousand")
   {
    result += temp;
    temp = 0;
   }
  }
  else
  {
   temp += str2int[number[i]];
  }
 }
 result += temp;
 printf("%d/n",result);
}

int main(int argc, char* argv[])
{
 Init();
 int count;
 scanf("%d",&count);
 getchar();
 while(count >= 1)
 {
  Input();
  Work();
  count--;
 }
 return 0;
}