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A. Strange Functions

2020/12/5 10:06:06 文章标签:

题目: Let’s define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321)123, f(120)21, f(1000000)1, f(111)111. Let’s define…

题目:
Let’s define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321)=123, f(120)=21, f(1000000)=1, f(111)=111.

Let’s define another function g(x)=xf(f(x)) (x is a positive integer as well).

Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1≤x≤n.

Input
The first line contains one integer t (1≤t≤100) — the number of test cases.

Each test case consists of one line containing one integer n (1≤n<10100). This integer is given without leading zeroes.

Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1,n].

Example
inputCopy
5
4
37
998244353
1000000007
12345678901337426966631415
outputCopy
1
2
9
10
26
Note
Explanations for the two first test cases of the example:

if n=4, then for every integer x such that 1≤x≤n, xf(f(x))=1;
if n=37, then for some integers x such that 1≤x≤n, xf(f(x))=1 (for example, if x=23, f(f(x))=23,xf(f(x))=1); and for other values of x, xf(f(x))=10 (for example, if x=30, f(f(x))=3, xf(f(x))=10). So, there are two different values of g(x).
题解:
简单的思维题,只需要输出数字的位数即可。
上代码!

#include <bits/stdc++.h>
using namespace std;
char a[300000];
int main()
{
	int t;
	cin>>t;
	while(t--){
		scanf("%s",a);
		int len=strlen(a);
		cout<<len<<endl;
	}
	return 0;
}

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