poj 1002 487-3279

poj 1002 487-3279

题意：给你n个字符串，均可以化简成为标准的字符串，问化简后的标准字符串的具体串，和每种的标准串的个数。

题解：将n个字符串化为全数字的字符串，进行字典序排序后，输出答案时加上需要的‘-’即可。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100010;

int p[26] = {2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8,9,9,9,9};
struct point
{
char a[8];
} ans
;
bool cmp(point x, point y)
{
return strcmp(x.a,y.a) < 0;
}
int main()
{
int n;
scanf("%d",&n);
for(int i = 0; i < n; i++)
{
char s[100];
scanf("%s",s);
int len = strlen(s);
int k = 0;
for(int j = 0; j < len; j++)
{
if(s[j] >= 'A' && s[j] <= 'Y')
ans[i].a[k++] = p[s[j]-'A']+'0';
else if(s[j] == '-')
continue;
else
ans[i].a[k++] = s[j];
}
ans[i].a[k] = '\0';
}
sort(ans,ans+n,cmp);
bool pan = 1;
int i = 0,j;
while(i < n)
{
j = i;
i++;
while(i < n && strcmp(ans[i].a,ans[j].a) == 0)
i++;
if(i - j > 1)
{
for(int k = 0; k <= 6; k++)
{
if(k == 3)
printf("-");
printf("%c",ans[j].a[k]);
}
printf(" %d\n",i-j);
pan = 0;
}
}
if(pan)
puts("No duplicates.");
return 0;
}