Xtreme9.0 - Digit Fun!

题目连接：

Description

An editorial, providing an approach to solve this problem, is presented at the bottom of this page.

Recurrence relations are an important tool for the computer scientist. Many algorithms, particularly those that use divide and conquer, have time complexities best modeled by recurrence relations. A recurrence relation allows us to recursively define a sequence of values by defining the nth value in terms of certain of its predecessors.

Many natural functions, such as factorials and the Fibonacci sequence, can easily be expressed as recurrences. The function of interest for this problem is described below.

Let |An| denote the number of digits in the decimal representation of An. Given any number A0, we define a sequence using the following recurrence:

Ai = |Ai-1| for i > 0

The goal of this problem is to determine the smallest positive i such that Ai = Ai-1.

Input

Input consists of multiple lines, each terminated by an end-of-line character. Each line (except the last) contains a value for A0, where each value is non-negative and no more than a million digits. The last line of input contains the word END.

Output

For each value of A0 given in the input, the program should output one line containing the smallest positive i such that Ai = Ai-1.

Sample Input

9999

0

1

9999999999

END

Sample Output

3

2

1

4

Hint

The first input value is A0 = 9999, resulting in A1 = |9999| = 4. Because 4 does not equal 9999, we find A2 = |A1| = |4| = 1. Since 1 is not equal to 4, we find A3 = |A2| = |1| = 1. A3 is equal to A2, making 3 the smallest positive i such that Ai = Ai-1.

The second input value is A0 = 0, resulting in A1 = |0| = 1. Because 0 does not equal 1, we find A2 = |A1| = |1| = 1. A2 is equal to A1, making 2 the smallest positive i such that Ai = Ai-1.

The third input value is A0 = 1, resulting in A1 = |1| = 1. A1 is equal to A0, making 1 the smallest positive i such that Ai = Ai-1.

The last input value is A0 = 9999999999, resulting in A1 = |9999999999| = 10. Because 10 does not equal 9999999999, we find A2 = |A1| = |10| = 2. Since 2 is not equal to 10, we find A3 = |A2| = |2| = 1. Since 1 is not equal to 2, we find A4 = |A3| = |1| = 1. A4 is equal to A3, making 4 the smallest positive i such that Ai = Ai-1.

题意

输入A[0]

定义F(x)=strlen(x)

A[i]=F(A[i-1])

求最小的i，使得A[i]=A[i-1]

题解

暴力水题

详细看代码

代码

#include
    
     using namespace std;int last;string s;string get(string ss){    string tmp;    int p=ss.size();    while(p)    {        tmp+=p%10+'0';        p/=10;    }    reverse(tmp.begin(),tmp.end());    return tmp;}int main(){    while(cin>>s)    {        if(s[0]=='E')break;//读到EOF时，break        last=0;        for(int i=0;i