【codechef】Superlucky Numbers（组合取模,枚举）

最新推荐文章于 2017-11-10 11:46:05 发布

原创最新推荐文章于 2017-11-10 11:46:05 发布 · 514 阅读

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本文介绍了一种算法，用于计算特定条件下由两个指定数字组成的n位超幸运数的数量，并通过模运算处理大数问题。文章详细展示了如何利用组合数学中的技巧解决这一问题。

Eshaan and Nischay decided to play a game. Nischay asked Eshaan to choose 2 distinct digits A and B. Eshaan calls a positive integer “lucky”, if it contains only the digits A and B. Nischay calls this number “superlucky” if sum of its digits is “lucky”.

For Example : If Eshaan chooses A = 1 and B = 3, then number 12 isn’t lucky but 133 or 1133131 are lucky. Also number 1133131 is superlucky but 133 is not.

Now Eshaan wants Nischay to find how many superlucky numbers of length exactly N are there.

Help Nischay to find the answer to this problem.

Since the answer can be very large, print answer modulo (10⁹+7).

Input

The first line of input contains T, denoting the number of test cases.
The second line of input contains 3 integers A,B and N.

Output

Print a single integer — the answer to the problem modulo (10⁹ +7).

Constraints

1 ≤ T ≤ 50
1 ≤ A < B ≤ 9
1 ≤ N ≤ 10⁶

Example

Input:
2
1 3 3
2 3 10

Output:
1
165

Explanation

Example case 1. Only one number of length 3 is superlucky which is 111 because 111 is lucky and sum of its digits 1+1+1=3 is also lucky.

http://www.codechef.com/problems/CLCO02

只由a和b组成且各个位上的和是lucky数且必须是n位的数字有几个？

#include<bits/stdc++.h>
using namespace std;
int a,b,n;
long long MOD=1e9+7;
long long fact[1000006]={0};
long long ifact[1000006]={0};

long long mulmod(long long a,long long b) //pow函数
 {	long long x=1,y=a;
 	while(b>0)
 	 {
 	 	if(b&1)
 	 	  x=(x*y)%MOD;
 	 	y=(y*y)%MOD;
 	 	b=b>>1;
 	 }
 	return x;
 }
 
bool isLucky(int m){
	while(m)
	 {
	 	if(m%10==a || m%10==b) // 这个数的某一位是a或b
	 	  m/=10;
	 	else 
	 	  return false;
	 }
	return true;
}

int main(){
	fact[0]=ifact[0]=1;
	for(int i=1;i<=1000000;i++)
	  fact[i]=(i*fact[i-1])%MOD;
	for(int i=1;i<=1000000;i++)
	  ifact[i]=mulmod(fact[i],MOD-2);
	
	int t;
	cin>>t;
	while(t--)
	 {
	 	cin>>a>>b>>n;
	 	int c=0;
	 	for(int i=0;i<=n;i++) //枚举这个数里面a出现的个数
	 	 {
	 	 	int m=i*a+b*(n-i); //各位上数字的和
	 	 	if(isLucky(m))   //判断这个和是否是lucky数
	 	 	  {
	 	 	  	long long ans=fact[n];
	 	 	  	long long den=(ifact[i]*ifact[n-i])%MOD; //组合模板，这些a摆放的方法数
	 	 	  	ans=(ans*den)%MOD;
	 	 	  	c=(c+ans)%MOD;
	 	 	  }
	 	 }
	 	 cout<<c<<endl;
	 }
}

下面是我总结的组合取模模板：

long long MOD=1e9+7;
long long fact[1000006];
long long ifact[1000006];
main(){
	fact[0]=ifact[0]=1;
	for(int i=1;i<=1000000;i++)
	  fact[i]=(i*fact[i-1])%MOD;
	for(int i=1;i<=1000000;i++)
	  ifact[i]=mulmod(fact[i],MOD-2);

	 int n,m;
	 cin>>n>>m;
	 long long ans=fact[n];
	 long long den=(ifact[m]*ifact[n-m])%MOD;
	 ans=(ans*den)%MOD;
	 cout<<ans<<endl;