For example, the string “acm” is mapped to 1 + 3 + 13 = 17. Unfortunately, this method does not give one-to-one mapping. The string “adl” also maps to 17 (1 + 4 + 12). This is called collision.
In this problem you will have to find the number of strings of length L, which maps to an integer S, using the above hash function. You have to consider strings that have only lowercase letters in strictly ascending order.
Suppose L = 3 and S = 10, there are 4 such strings.
- abg
- acf
- ade
- bce
agb also produces 10 but the letters are not strictly in ascending order.
bh also produces 10 but it has 2 letters.
Input
There will be several cases. Each case consists of 2 integers L and S. (0 < L, S < 10000). Input is terminated with 2 zeros.
Output
For each case, output Case #: where # is replaced by case number. Then output the result. Follow the sample for exact format. The result will fit in 32 bit signed integers.
| Sample Input | Output for Sample Input |
| 3 10 2 3 0 0 | Case 1: 4 Case 2: 1 |
#include<stdio.h>
int main()
{
int fun(int,int,int,int,int,int,int[]);
int t,inp,num,A[10000]={0};
scanf("%d %d",&inp,&num);
if(inp==0 && num==0)
{
return 0;
}
else
{
printf("Count= %d\n",fun(1,inp,0,0,num,0,A));
}
return 0;
}
int fun(int level,int inp,int sum,int curr,int num,int count,int A[10000])
{
if(level==inp)
{
if(num-sum<=26 && num-sum>curr)
{
for(int i=0;i<level-1;i++)
{
printf("%d ",A[i]);
}
printf("%d\n",num-sum);
count+=1;
return count;
}
else
{
return count;
}
}
else
{
for(int i=curr+1;i<=(num/(inp-1));i++)
{
sum+=i;
A[level-1]=i;
count=fun(level+1,inp,sum,i,num,count,A);
sum-=i;
}
return count;
}
}
