11609  Teams
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11609  Teams
The problem says that output has to be printed result modulo 1000000007, I dont understand why this value and moreover for n=3, the number of ways the coach can choose his team is:
First for k = 1, He has 3 ways,
for k = 2, He has 6 * 2 (For different captains) ways,
for k = 3, He has 1 * 3 (For different captains) ways,
So totally he has 3+12+3 = 18 ways, then 18 modulo 1000000007 is 18 itself. I don't understand how the output is given as 1.
Let me know if I have understood the problem correctly.
First for k = 1, He has 3 ways,
for k = 2, He has 6 * 2 (For different captains) ways,
for k = 3, He has 1 * 3 (For different captains) ways,
So totally he has 3+12+3 = 18 ways, then 18 modulo 1000000007 is 18 itself. I don't understand how the output is given as 1.
Let me know if I have understood the problem correctly.

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Re: 11609  Teams
The first line of the input represents the number of cases in the input.
11609 ' Why WA !!
Code: Select all
#include<stdio.h>
#include<math.h>
#define mod 1000000007
int main()
{
int n,i;
unsigned long m,p;
scanf("%d",&n);
for(i=1; i<=n; i++)
{
scanf("%lu",&m);
p=pow(2,(m1));
printf("Case #%d: %lu\n",i,(m*p)%mod);
}
return 0;
}
Last edited by Psh on Thu Dec 01, 2011 2:53 pm, edited 1 time in total.
Re: 11609 ' Why WA !!
What's your output for the input > 1000000000
pow(2,(10000000001)) .. this statement won't work due to overflow.
pow(2,(10000000001)) .. this statement won't work due to overflow.

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Re: 11609 ' Why WA !!
Can U imagine what is the value of pow(2,m1) where m=1000000000. Value of pow(2,31) is 2147483648.So i think you
can understand what pow(2,999999999) will return. So you have to write your own power function
where the intermediate result is mod(%) by 1000000007. And you can't avoid TLE by iteratively
calculating power, like using a for loop till N.There is a nice divide and conquer algorithm for calculating (b^p)%m ,where
b is base,p is power and m is mod.Try this.
can understand what pow(2,999999999) will return. So you have to write your own power function
where the intermediate result is mod(%) by 1000000007. And you can't avoid TLE by iteratively
calculating power, like using a for loop till N.There is a nice divide and conquer algorithm for calculating (b^p)%m ,where
b is base,p is power and m is mod.Try this.
Last edited by sazzadcsedu on Wed Jul 07, 2010 7:01 pm, edited 1 time in total.
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Re: 11609  Teams
I think, the answer is: summation of k * nCk for k = 1 ... n [nCk = n Choose k]
Can someone tell me how to get a closed form?
Can someone tell me how to get a closed form?
regards,
nymo
nymo
Re: 11609  Teams
It's easy to derive this summation formula...after deriving this, why don't you write down all the values of this summation from n = 1 to n = 7 on paper? May be you will notice a pattern. If you do notice a pattern may be you will understand why and how such a pattern exists. Good luck.
You tried your best and you failed miserably. The lesson is 'never try'. Homer Simpson

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Re: 11609  Teams
I am not getting the required output: I m using the summation of k*nCk
Code: Select all
#include<cstdio>
using namespace std;
#define mod 1000000007;
int main()
{
int n,i,j,c=0,num,k;
long long int sum,res;
scanf("%d",&n);
while(n)
{
c++;
sum=0;
scanf("%d",&k);
for(i=1;i<=k;i++)
{
res=i;
num=k;
for(j=1;j<=i;j++)
{
res=res*num;
res=res/j;
num;
res=res%mod;
}
sum=sum+res;
sum=sum%mod;
//printf("%d %lld %lld\n",i,sum,res);
}
printf("Case #%d: %lld\n",c,sum);
}
return 0;
}

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Re: 11609  Teams
Your code is getting TLE. Try input:
Code: Select all
1
1000000000
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