Sdut 2165 Crack Mathmen(数论）(山东省ACM第二届省赛E 题)

Crack Mathmen

TimeLimit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

 

题目描写叙述

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

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

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

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

3. the third character is 'r', and it'sASCII 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'sASCII code is 108. Then f(′l′) =108^2 mod 997 = 697.

5. the fifth character is 'd', and it'sASCII 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 mathmansent was intercepted by the human being. As the cleverest one, could youfind out what the plain text (i.e., the message before encryption) was?

输入

 The input contains multiple test cases. The first line ofthe input contains a integer, indicating the number of test cases in theinput. 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 lengthof the string is at most 10^6.

输出

 For each test case, output a line containing the plaintext. If their are no or more than one possible plain text that can be encryptedas 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.

演示样例输入

3

2

590357035697030

0

001001001001001

1000000000

001001001001001

演示样例输出

World

No Solution

No Solution

/*************************

一道非常高大上的数论题，開始看的一道  大神的，用数论的方法解的：http://limyao.com/?p=113#comment-111

大神用的有素数原根，全然剩余系，离散对数，模线性方程，知识点非常多，也非常难。。

有点小困难，然后我和小伙伴修昊讨论了下，认为最初的想法——打表应该能够，然后就付诸行动了。。

我写的时候有一点没想通，也是非常关键的一点，加密算法  原码 转换到  加password，加password会出现反复的情况，这个我没推断到，后开在小伙伴的解释下，瞬间顿悟，然后，恩就A了。

**********************/

Code:

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
char ar[1000],br[340000],str[10000005];
int cal(int temp,int t)//位运算高速幂模
{
    int ans = 1;
    while(t)
    {
        if(t&1)
            ans = (ans * temp) % 997;
        temp = temp * temp % 997;
        t = t >> 1;
    }
    return ans;
}


bool init(int n)
{
    memset(ar,'\0',sizeof(ar));
    int i,tmp;
    for(i = 32;i<=126;i++)  //  ASCII 码 打表，   
    {
        if(ar[cal(i,n)]=='\0')    //  推断 原码 ->加password 转换过程中是否反复，反复则直接返回false
            ar[cal(i,n)] = char(i);  //  加password 作数组下标，匹配时直接寻找，无需查找
        else
            return false;
    }
    return true;
}

int main()
{
    int n,c,i,j,len,cur;
    bool now;
    cin>>c;
    while(c--)
    {
        now = true;
        memset(br,'\0',sizeof(br));
        cin>>n;
        cin>>str;
        len = strlen(str);
        j = 0;
        if(init(n))
        {
            for(i = 0;i<len;i+=3)
            {
                cur = (str[i]-'0') * 100 + (str[i+1]-'0') * 10 + str[i+2] - '0';
                if(ar[cur] == '\0')
                {
                    now = false;
                    break;
                }
                br[j++] = ar[cur];
            }
        }
        else
            now = false;
        if(n==0)
            now = false;  //  n = 0 时 肯定为 No Solution
        if(now)
            cout<<br<<endl;
        else
            cout<<"No Solution"<<endl;
    }
    return 0;
}