ACM-ICPC 2018 徐州赛区网络预赛 I

本文介绍了一种新颖的字符串加密算法,该算法通过选取一个字符作为种子,计算输入字符串中每个字符与种子之间的绝对值差,并将结果转换为固定长度的数字串。文章详细解释了算法的工作原理,并提供了一个示例,展示了如何使用该算法处理特定的输入字符串。

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Characters with Hash

Mur loves hash algorithm, and he sometimes encrypt another one's name, and call him with that encrypted value. For instance, he calls Kimura KMR, and calls Suzuki YJSNPI. One day he read a book about SHA-256256 , which can transit a string into just 256256 bits. Mur thought that is really cool, and he came up with a new algorithm to do the similar work. The algorithm works this way: first we choose a single letter L as the seed, and for the input(you can regard the input as a string ss, s[i]s[i] represents the iith character in the string) we calculates the value(|(int) L - s[i]|∣(int)L−s[i]∣), and write down the number(keeping leading zero. The length of each answer equals to 22 because the string only contains letters and numbers). Numbers writes from left to right, finally transfer all digits into a single integer(without leading zero(ss)). For instance, if we choose 'z' as the seed, the string "oMl" becomes "1111 4545 1414".

It's easy to find out that the algorithm cannot transfer any input string into the same length. Though in despair, Mur still wants to know the length of the answer the algorithm produces. Due to the silliness of Mur, he can even not figure out this, so you are assigned with the work to calculate the answer.

Input

First line a integer TT , the number of test cases (T \le 10)(T≤10).

For each test case:

First line contains a integer NN and a character zz, (N \le 1000000)(N≤1000000).

Second line contains a string with length NN . Problem makes sure that all characters referred in the problem are only letters.

Output

A single number which gives the answer.

样例输入复制

2
3 z
oMl
6 Y
YJSNPI

样例输出复制

6
10

答案:

Java:

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int num = sc.nextInt();
		int sss[] = new int[num];
		int n;
		char c;
		String s;
		
		for (int i = 0; i < num; i++) {
			n = sc.nextInt();
			c = sc.next().charAt(0);
			s = sc.next();
			
			sss[i] = f(n,c,s);
		}
		
		for (int i = 0; i < sss.length; i++) {
			System.out.println(sss[i]);
		}
	}

	private static int f(int n, char c, String s) {
		int tol = s.length()*2;
		for (int i = 0; i < n; i++) {
			int temp = c - s.charAt(i);
			if(temp==0) {
				tol -= 2;
			}else if(temp<10&&temp>-10) {
				tol -=1;
				break;
			}else {
				break;
			}
		}
		if(tol==0) {
			tol = 1;
		}
		
		return tol;
	}
}

c:

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MOD = (int)1e9 + 7;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int maxn = (int)1e6 + 10;

int n;
char s[10];
char str[maxn];

int arr[maxn];

inline void RUN()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        scanf("%s", s);
        scanf("%s", str);
        int len = strlen(str);
        for (int i = 0; i < len; ++i)
        {
            arr[i] = abs(str[i] - s[0]);
        }
        int ans = 2 * len;
        for (int i = 0; i < len; ++i)
        {
            if (arr[i] == 0)
            {
                ans -= 2;
            }
            else if (arr[i] < 10)
            {
                ans -= 1;
                break;
            }
            else
            {
                break;
            }
        }
        if (ans == 0) ans = 1;
        printf("%d\n", ans);
    }
}

int main()
{
#ifdef LOCAL_JUDGE
    freopen("Text.txt", "r", stdin);
#endif // LOCAL_JUDGE

    RUN();

#ifdef LOCAL_JUDGE
    fclose(stdin);
#endif // LOCAL_JUDGE
}

 

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