【刷题】Sum of Digits【1】

司六米希

发布于 2022-11-15 19:22:21

3230

发布于 2022-11-15 19:22:21

文章被收录于专栏：司六米希司六米希

【刷题】Sum of Digits【1】

一、题目

1.题目描述

题目：

输入：

输出：

示例 :

提示: In the first sample the number already is one-digit — Herald can’t cast a spell. The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once. The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit.

二、解题报告

1.思路分析

不断对数字按位进行累加，直到变成一个数字为止

2.代码详解

python👇【检验超时】

a=input()
b=list(a)
c=[]
s=0
i=1
while 1:
    for j in range(len(b)):
        s+=int(b[j])
    if s==0:
        print(0)
        break
    if s<9:
        print(i)
        break
    else:
        b=list(str(s))
        c=[]
        s=0
        i+=1

C++👇

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include<set>
#include <algorithm>
using namespace std;
 
char s[100100];
 
int main() 
{
	int n,cnt,i,res;
	scanf("%s",s);
	cnt=0;
	while (s[1]) 
	{
		cnt++;
		res=0;
		for(i=0;s[i];i++)
		{
			res+=s[i]-'0';
		}
		sprintf(s,"%d",res);
	}
	printf("%d\n",cnt);
	return 0;
}

本文参与腾讯云自媒体分享计划，分享自作者个人站点/博客。

原始发表：2022-06-30，如有侵权请联系 cloudcommunity@tencent.com 删除

c++