POJ 1002 487-3279

1. 试了各种数据类型保存输入的字符串，比如string，字符数组。在string类型中，不能一个一个字符地拷贝字符串，因为字符串末尾还有一个字符。

2. 字符型数字转换成int类型；

3. 自我感觉change函数写的比较巧妙，跟Curling
2.0有点类似。是把一个数组（有重复元素）无重复的放到另外一个数组。

#include<stdio.h>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
using namespace std;
int telnum[100010];
int tel[100010][7];
int n;
struct Node
{
int ans;
int num;
} node[100010];
bool cmp(int a, int b)
{
return a < b;
}
void change(int k, string str)
{
int i, j = 0;
for(i = 0; i < 7; i++)
{

while(str[j] == '-')
j++;
if(str[j] == '0' || str[j] == '1' || str[j] == '2' || str[j] == '3' || str[j] == '4' || str[j] == '5' || str[j] == '6' || str[j] == '7' || str[j] == '8' || str[j] == '9')
{
tel[k][i] = str[j] - '0';//操，=写成==了，测试了一个多小时。。。
j++;
continue;
}
if(str[j] == 'A' || str[j] == 'B' || str[j] == 'C')
{
tel[k][i] = 2;
j++;
continue;
}

if(str[j] == 'D' || str[j] == 'E' || str[j] == 'F')
{
tel[k][i] = 3;
j++;
continue;
}
if(str[j] == 'G' || str[j] == 'H' || str[j] == 'I')
{
tel[k][i] = 4;
j++;
continue;
}
if(str[j] == 'J' || str[j] == 'K' || str[j] == 'L')
{
tel[k][i] = 5;
j++;
continue;
}
if(str[j] == 'M' || str[j] == 'N' || str[j] == 'O')
{
tel[k][i] = 6;
j++;
continue;
}
if(str[j] == 'P' || str[j] == 'R' || str[j] == 'S')
{
tel[k][i] = 7;
j++;
continue;
}
if(str[j] == 'T' || str[j] == 'U' || str[j] == 'V')
{
tel[k][i] = 8;
j++;
continue;
}
if(str[j] == 'W' || str[j] == 'X' || str[j] == 'Y')
{
tel[k][i] = 9;
j++;
continue;
}
}
}
int getnum()
{
int i, j = 0;
for(i = 0; j < n; i++)
{
node[i].num = 0;
node[i].ans = telnum[j];
node[i].num++;
while(telnum[j] == telnum[j+1])
{
j++;
node[i].num++;
}
j++;
}
return i;
}

int main()
{
string str;
int count = 0;
cin >> n;
memset(tel, 0, sizeof(tel));
for(int i = 0; i < n; i++)
{
cin >> str;
change(i, str);
/*        for(int j = 0; j < 7; j++)
{
cout << tel[i][j];
}
cout << endl;*/
telnum[i] = tel[i][0] * 1000000 + tel[i][1] * 100000 + tel[i][2] * 10000 + tel[i][3] * 1000 + tel[i][4] * 100 + tel[i][5] * 10 + tel[i][6];
}
sort(telnum, telnum + n, cmp);
int m = getnum();
for(int i = 0; i < m; i++)
{
if(node[i].num > 1)
{
cout << node[i].ans / 10000 << '-' << node[i].ans % 10000 << ' ' << node[i].num << endl;
}
else count++;
}
if(count == n)
cout << "No duplicates." << endl;
return 0;
}