Turing Tape CodeForces

codeforces 133c

题意

给你一串字符串，包含1~105个字符。按照规则输出数字。

1. The 8-bit binary notation of the ASCII-code of the previous printed character is reversed. When the first element of the array is processed, the result of this step is considered to be 0. 将ASCII码反转过来。

2. The i-th element of the array is subtracted from the result of the previous step modulo 256. 用前一个值减去这个值，并且模256，如果这是第一个那么前一个值为0.

3. The binary notation of the result of the previous step is reversed again to produce ASCII-code of the i-th character to be printed. 输出，继续。

样例

Input

Hello, World!

Output

238
108
112
0
64
194
48
26
244
168
24
16
162

Note

Let's have a closer look at the beginning of the example. The first character is "H" with ASCII-code 72 = 010010002. Its reverse is 000100102 = 18, and this number should become the result of the second step of processing. The result of the first step is considered to be 0, so the first element of the array has to be (0 - 18) mod 256 = 238, where a mod b is the remainder of division of a by b.

思路

• 按照他的步骤做就好了
• 输入的时候卡了一下，要用getline(cin,s); s是string类型的。

代码

#include<bits/stdc++.h>
using namespace std;
const int MAXN=105;
int a[MAXN];
const int MOD= 256;
string s;
int main(){
	getline(cin,s);
	int len=s.size();
	int tmp,t;
	for(int i=1;i<=len;i++){
		a[i]=s[i-1];
		tmp=a[i];
		t=0;
		int j=0;
		while(tmp){
			j++;
			if(tmp&1){
				t+=1<<(8-j);
			}
			tmp>>=1;
		}
		a[i]=t;
		printf("%d
", (a[i-1]-a[i]+MOD)%MOD);
	}	
	return 0;
}