Sum of you

• Sum of you .

The time limit ：1 second
Space restriction ：128M

Title Description

Small brother is super powerful , Any number can be changed into the sum of numbers on its various digits through one operation .

Now little brother has found a number n, After several operations, it can become a single digit .
​

Input description

Input contains an integer $n$

Data range

$1<=n<=10^{100000}$

Output description

Output a number per line to indicate the answer

Example 1

Input

991

Output

3

Topic analysis

Directly as required , Just simulate .

In fact, I see $10^{100000}$ Don't be afraid , Directly treat the input number as a string .

$10^{100000}$ That is, the maximum length of the input string is $100000$, and $100000$ The maximum sum of numbers is less than $10^6$（ most $6$ position ）

So it will soon converge to a single digit . Don't worry about time at all .

character string の Everyone and

The bit sum of the string is relatively easy , Traversing the string directly , And add up every bit .

ll getS(string s) {
    
    ll ans = 0;
    for (char& c : s)
        ans += c - '0';
    return ans;
}

Numbers の Everyone and

The sum of figures is not difficult . When the number is not zero , Take out the single digit of the number every time , Then divide the number by $10$ that will do .

ll getS(ll n) {
    
    ll ans = 0;
    while (n) {
    
        ans += n % 10;
        n /= 10;
    }
    return ans;
}

AC Code

/* * @Author: LetMeFly * @Date: 2022-07-21 09:46:18 * @LastEditors: LetMeFly * @LastEditTime: 2022-07-21 09:48:26 */
#include <bits/stdc++.h>
using namespace std;
#define mem(a) memset(a, 0, sizeof(a))
#define dbg(x) cout << #x << " = " << x << endl
#define fi(i, l, r) for (int i = l; i < r; i++)
#define cd(a) scanf("%d", &a)
typedef long long ll;

ll getS(string s) {
    
    ll ans = 0;
    for (char& c : s)
        ans += c - '0';
    return ans;
}

ll getS(ll n) {
    
    ll ans = 0;
    while (n) {
    
        ans += n % 10;
        n /= 10;
    }
    return ans;
}

int main() {
    
    string s;
    cin >> s;
    if (s.size() == 1) {
    
        puts("0");
        return 0;
    }
    ll n = getS(s);
    int ans = 1;
    while (true) {
    
        if (n < 10)
            break;
        ans++;
        n = getS(n);
    }
    cout << ans << endl;
    return 0;
}

Although the code can be copied , But it's better to knock after understanding

Originality is not easy. , Reprint please attach Link to the original text Oh ~
Tisfy：https://letmefly.blog.csdn.net/article/details/125918169