SDUT 2165 Crack Mathmen（快速幂）山东省第二届ACM大学生程序设计竞赛

Crack Mathmen

Time Limit: 1000MS Memory limit: 65536K

题目描述

 Since mathmen take security very seriously, they communicate in encrypted messages. They cipher their texts in this way: for every characther c in the message, they replace c with f(c) = (the ASCII code of c)n mod 1997 if f(c) < 10, they put two preceding zeros in front of f(c) to make it a three digit number; if 10 <= f(c) < 100, they put one preceding zero in front of f(c) to make it a three digit number.

For example, if they choose n = 2 and the message is "World" (without quotation marks), they encode the message like this:

1. the first character is 'W', and it's ASCII code is 87. Then f(′W′) = 87^2 mod 997 = 590.

2. the second character is 'o', and it's ASCII code is 111. Then f(′o′) = 111^2 mod 997 = 357.

3. the third character is 'r', and it's ASCII code is 114. Then f(′r′) = 114^2 mod 997 = 35. Since 10 <= f(′r′) < 100, they add a 0 in front and make it 035.

4. the forth character is 'l', and it's ASCII code is 108. Then f(′l′) = 108^2 mod 997 = 697.

5. the fifth character is 'd', and it's ASCII code is 100. Then f(′d′) = 100^2 mod 997 = 30. Since 10 <= f(′d′) < 100, they add a 0 in front and make it 030.

6. Hence, the encrypted message is "590357035697030".

One day, an encrypted message a mathman sent was intercepted by the human being. As the cleverest one, could you find out what the plain text (i.e., the message before encryption) was?

输入

 The input contains multiple test cases. The first line of the input contains a integer, indicating the number of test cases in the input. The first line of each test case contains a non-negative integer n (n <= 10^9). The second line of each test case contains a string of digits. The length of the string is at most 10^6.

输出

 For each test case, output a line containing the plain text. If their are no or more than one possible plain text that can be encrypted as the input, then output "No Solution" (without quotation marks). Since mathmen use only alphebetical letters and digits, you can assume that no characters other than alphebetical letters and digits may occur in the plain text. Print a line between two test cases.

示例输入

3
2
590357035697030
0
001001001001001
1000000000
001001001001001

示例输出

World
No Solution
No Solution

分析：题意是，对字母和数字进行编码，每个字母或数字的编码都是三位，编码规则为该字符的ASCII值的n次方mod 997。并保证对每个字符（只包括字母和数字）的编码存在且唯一，若译码结果不是字符或数字则输出No Solution

用快速幂求出相应的编码，再用编码做为数字下表存字符，例如ch[ 590 ]='W';

代码如下：

	#include <stdio.h>
	#include <string.h>
	long long powerMod(long long a,long long b,long long c)
	{
		long long ans=1;
		a=a%c;
		while(b>0)
		{
			if(b%2==1)
				ans=(ans*a)%c;
			b=b/2;
			a=(a*a)%c;
		}
		return ans;
	}
	
	int main()
	{
		int i,j;
		int T;
		int b,c;
		long long n;
		char a[1005],str[1000005];
		scanf("%d",&T);
		while(T--)
		{
			int peace=0;
			scanf("%I64d%*c",&n);
			memset(a,' ',sizeof(a));
			for(i=0;i<26;i++)
			{
				long long ans=powerMod('a'+i,n,997);<span style="white-space:pre">	</span>//快速幂小写字母编码
				if(a[ans]==' ')<span style="white-space:pre">		</span>//保证编码不重复
					a[ans]='a'+i;
				else
				{
					peace=1;
					break;
				}
			}
			
			for(i=0;i<26 && !peace;i++)
			{
				long long ans=powerMod('A'+i,n,997);<span style="white-space:pre">	</span>//快速幂大写字母编码
				if(a[ans]==' ')
					a[ans]='A'+i;
				else
				{
					peace=1;
					break;
				}
			}
			
			for(i=0;i<10 && !peace;i++)
			{
				long long ans=powerMod('0'+i,n,997);<span style="white-space:pre">	</span>//快速幂数字编码
				if(a[ans]==' ')
					a[ans]='0'+i;
				else
				{
					peace=1;
					break;
				}
			}
			scanf("%s",str);
			if(!peace)
			{
					int len=strlen(str);
					for(i=0;i<len;i+=3)
					{
						int ans;
						ans=(str[i]-'0')*100+(str[i+1]-'0')*10+(str[i+2]-'0');
						if(a[ans]!=' ')
							printf("%c",a[ans]);
						else
						{
							printf("No Solution");
							break;
						}
					}
					putchar('\n');
			}
			else
			{
				printf("No Solution\n");
			}
		}	
		return 0;
	}