Description
It's easy to tell if two words are identical - just check the letters. But how do you tell if two words are almost identical?
And how close is "almost"?
There are lots of techniques for approximate word matching. One is to determine the best substring match, which is the number of common letters when the words are compared letter-byletter.
The key to this approach is that the words can overlap in any way. For example, consider the words CAPILLARY and MARSUPIAL. One way to compare them is to overlay them:
CAPILLARY
MARSUPIAL
There is only one common letter (A). Better is the following overlay:
CAPILLARY
MARSUPIAL
with two common letters (A and R), but the best is:
CAPILLARY
MARSUPIAL
Which has three common letters (P, I and L).
The approximation measure appx(word1, word2) for two words is given by:
common letters * 2
-----------------------------
length(word1) + length(word2)
Thus, for this example, appx(CAPILLARY, MARSUPIAL) = 6 / (9 + 9) = 1/3. Obviously, for any word W appx(W, W) = 1, which is a nice property, while words with no common letters have an appx value of 0.
Input
The input for your program will be a series of words, two per line, until the end-of-file flag of -1.
Using the above technique, you are to calculate appx() for the pair of words on the line and print the result.
The words will all be uppercase.
Output
Print the value for appx() for each pair as a reduced fraction,Fractions reducing to zero or one should have no denominator.
Sample Input
CAR CART
TURKEY CHICKEN
MONEY POVERTY
ROUGH PESKY
A A
-1
Sample Output
appx(CAR,CART) = 6/7
appx(TURKEY,CHICKEN) = 4/13
appx(MONEY,POVERTY) = 1/3
appx(ROUGH,PESKY) = 0
appx(A,A) = 1
*/
#include<stdio.h>
#include<string.h>
int gcd(int m,int n)//求最大公约数；
{
if(n==0)
return m;
else
return gcd(n,m%n);
}
int main()
{
char a[100],b[100];
while(scanf("%s",a)!=EOF)
{
if(strcmp(a,"-1")==0)
break;
else
scanf("%s",b);
int i,j,k,l,max=0,t;
int len1,len2,len;
len1=strlen(a);
len2=strlen(b);
for(i=0;i<len1;i++)/*相当于a[len1]不动，从,i=j=0開始，b[i++]与a[j++]比較。同样的话t++，
之后b[0]与a[i]比較，至到a[len-1]与b[i]比較记下t，并与之前的t比較,得出更大的t。后面继续从b[1]继续比較，直到最好能比較结束*/
{
k=0;
for(l=i,j=0;l<len1;l++,j++)
{
if(a[l]==b[j])
k++;
}
max=max>k?
max:k;
}
for(i=0;i<len2;i++)//相当于b[len2]不动，与上面类似，比較easy举一反三。
{
k=0;
for(l=i,j=0;l<len2;l++,j++)
{
if(b[l]==a[j])
k++;
}
max=max>k?
max:k;
}
len=len1+len2,max*=2;
t=gcd(max,len);
if(len==max)
{
printf("appx(%s,%s) = 1
",a,b);
continue;//这个continue不能省略；
}
if(max==0)
{
printf("appx(%s,%s) = 0
",a,b);
continue;
}
else
printf("appx(%s,%s) = %d/%d
",a,b,max/t,len/t);
}
return 0;
}