第十四届华中科技大学程序设计竞赛（Beautiful Trees Cutting）二项式定理 & 快速幂 & 费马小定理求逆元

It’s universally acknowledged that there’re innumerable trees in the campus of HUST.

One day Xiao Ming is walking on a straight road and sees many trees line up in the right side. Heights of each tree which is denoted by a non-negative integer from 0 to 9 can form a tree string. It's very surprising to find that the tree string can be represent as an initial string repeating K times. Now he wants to remove some trees to make the rest of the tree string looks beautiful. A string is beautiful if and only if the integer it represents is divisible by five. Now he wonders how many ways there are that he could make it.

Note that when we transfer the string to a integer, we ignore the leading zeros. For example, the string “00055” will be seen as 55. And also pay attention that two ways are considered different if the removed trees are different. The result can be very large, so printing the answer (mod 1000000007) is enough. And additionally, Xiao Ming can't cut down all trees.

输入描述:

The first line contains a single integer K , which indicates that the tree string is the initial string repeating K times.

The second line is the initial string S .

输出描述:

A single integer, the number of ways to remove trees mod 1000000007.

示例1

输入

1
100

输出

6

说明

Initially, the sequence is ‘100’. There are
6 ways:
100
1_0
10_
_00
__0
_0_

示例2

输入

3
125390

输出

149796

题意：输入一个看k，表示 下面字符串要重复多少次；下面一行为字符串，在重复k次的字符串中去掉一些字符，使得能整除5，去掉字符不同，则为不同的方法，最后结果可能会很大，mod  1000000007

思路：能乘除5的一定是0或者5结尾，所以 只要当前字符是0或者5的话，那么就可以把它之前的 字符去掉任意改个；若当前字符是i的话，前面有i个字符，那么 从前面字符 选出 0 去掉，从中选出 1 去掉，从中选出 2去掉......从中选出 i个去掉

求逆元的解释： 点击打开链接

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define Max 100010
#define ll long long
const long long mod = 1000000007;
char str[Max];
ll p_pow(ll a,ll b)    // a^b 
{
	ll res = 1;
	while(b)
	{
		if(b&1) res = res*a%mod;
			b>>=1;
		a = a*a % mod; 
	}
	return res;
}
int main()
{
	ll i,j,k;
	while(~scanf("%lld",&k))
	{
		scanf("%s",str);
		ll l = strlen(str);
		ll q = p_pow(2,l);
		ll fz = p_pow(q,k);
		fz = ((1-fz)%mod + mod)%mod;
		ll fm = ((1-q)%mod+mod)%mod;
		ll temp = fz*(p_pow(fm,mod-2))%mod;    // 费马小定理求逆元； 
		ll res = 0;
		for(i = 0;i<l;i++)
		{
			if(str[i] == '5'||str[i]=='0')
			{
				//res += p_pow(2,i)*temp%mod;  // 我不知道这样写是不是溢出了，但是他就是不对了；
				res +=p_pow(2,i);  // 把 temp 提出来，先让括号内的相加； 
			}
		}
		res = (res%mod)*temp%mod;   
		printf("%lld\n",res);
	}
	return 0;
}