山东省第二届ACM大学生程序设计竞赛 Crack Mathmen 打表 模拟

Crack Mathmen

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

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

3259035703569703000010010010010011000000000001001001001001

示例输出

WorldNo SolutionNo Solution

提示

 

来源

 山东省第二届ACM大学生程序设计竞赛

给你加密的方式

我们可以打表处理所有的结果然后输出就行

一开始以为只有对字母加密然后无限wa

ACcode:

#include <iostream>#include <cstdio>#include <cstring>#include <string>#define maxn 10000005#define mod 997#define ll long longusing namespace std;char s[maxn];char has[1000];char ans[1234567];bool flag;int pow_mod(int x,int n){int res=1;x=x%mod;while(n>0){if(n%2)res=res*x%mod;x=x*x%mod;n/=2;}return res;}bool init(int n){    memset(has,'\0',sizeof(has));    char aaa;    for(int i=32;i<=126;++i){        aaa=i;        int id=pow_mod(i,n);        if(has[id]!='\0')return  false;        has[id]=aaa;    }    return true;}int main(){    int loop,n;    scanf("%d",&loop);    while(loop--){        scanf("%d",&n);        scanf("%s",s);        flag=true;        flag=init(n);        if(!n)flag=false;        int tot=0;        if(flag){            int len=strlen(s);            int tmp;            for(int i=0;i<len;i+=3){                tmp=(s[i]-'0') * 100+(s[i+1]-'0') * 10+s[i+2] - '0';                if(has[tmp]!='\0')ans[tot++]=has[tmp];                else {                    flag=false;                    break;                }            }            ans[tot]=0;        }        printf("%s\n",flag?ans:"No Solution");    }    return 0;}/*42590357035697030000100100100100110000000000010010010010012590357035697030*/