摩托羅拉Moto G 2015款釋出：IPX7級防水

2015-07-28 22:22:02

IT之家訊 7月28日消息，經曆過頻繁曝光後，摩托羅拉今晚正式釋出了2015款Moto G，雖然之前的曝光已經将這款産品介紹的七七八八，但在這裡我們還是要“光明正大”的重新審視下這款手機。

與之前曝光的配置幾乎一緻，Moto G 2015采用了5英寸720P顯示屏，搭載64位四核骁龍410處理器，前置500萬像素攝像頭+後置1300萬像素攝像頭，電池容量為2470mAh電池。

此外，Moto G 2015還分為兩個不同的版本，分别是1GB RAM+8GB ROM版和2GB RAM+16GB ROM版，支援記憶體卡擴充、支援4G LTE。該機最大的亮點是具備IPX7級别防水功能。可以在一米深的水中“存活”半小時左右。

目前該機的印度版本售價已經公布，1GB RAM版售價11999盧比（約合人民币1168元），2GB RAM版售價12999盧比（約合人民币1265元）。8GB版的美國售價為179.99美元（約合人民币1117元）。據悉，摩托羅拉今晚一共推出了三款手機，分别是Moto G 2015款、Moto X Style、Moto X Play。

相關閱讀：《摩托羅拉Moto X Style釋出：2100萬像素攝像頭+2K屏》

1024. Palindromic Number (25)（C++)

2023-04-14 16:11:13

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

Non-palindromic numbers can be paired with palindromic ones via a series of operations. First, the non-palindromic number is reversed and the result is added to the original number. If the result is not a palindromic number, this is repeated until it gives a palindromic number. For example, if we start from 67, we can obtain a palindromic number in 2 steps: 67 + 76 = 143, and 143 + 341 = 484.

Given any positive integer N, you are supposed to find its paired palindromic number and the number of steps taken to find it.

Input Specification:

Each input file contains one test case. Each case consists of two positive numbers N and K, where N (<= 1010) is the initial numer and K (<= 100) is the maximum number of steps. The numbers are separated by a space.

Output Specification:

For each test case, output two numbers, one in each line. The first number is the paired palindromic number of N, and the second number is the number of steps taken to find the palindromic number. If the palindromic number is not found after K steps, just output the number obtained at the Kth step and K instead.

Sample Input 1:

67 3

Sample Output 1:

484
2

Sample Input 2:

69 3

Sample Output 2:

1353 3

知识点：reverse(s.begin(),s.end) //头文件在algorithm 字符串反转。

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
string s, ans;
int k, temp, c = 0, cnt = 0;
int main(){
	cin >> s >> k;
	ans = s;
	reverse(ans.begin(),ans.end());
	for(; cnt < k; ++ cnt){
		if(s == ans)
			break;
		ans = ""; c = 0;
		for(int i = s.length() - 1; i >= 0; -- i){
			temp = s[i] + s[s.length() - i - 1] - '0' - '0' + c;
			ans += (temp % 10) + '0';
			c = temp / 10;
		}
		if(c == 1)
			ans += '1';
		s = ans;
		reverse(s.begin(), s.end());
	}
	printf("%s\n%d", s.c_str(), cnt);
}

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