POJ2065 TOJ1028 ：SETI 高斯消元
描述
For some years, quite a lot of work has been put into listening to electromagnetic radio signals received from space, in order to understand what civilizations in distant galaxies might be trying to tell us. One signal source that has been of particular interest to the scientists at Universit?Le de Technologie Spatiale is the Nebula Stupidicus.
Recently, it was discovered that if each message is assumed to be transmitted as a sequence of integers a0, a1, ...an-1 the function
These relationships altogether have been considered too peculiar for being pure coincidences, which calls for further investigation.
The linguists at the faculty of Langues et Cultures Extraterrestres transcribe these messages to strings in the English alphabet to make the messages easier to handle while trying to interpret their meanings. The transcription procedure simply assigns the letters a..z to the values 1..26 that f (k) might evaluate to, such that 1 = a, 2 = b etc. The value 0 is transcribed to '*' (an asterisk). While transcribing messages, the linguists simply loop from k = 1 to n, and append the character corresponding to the value of f (k) at the end of the string.
The backward transcription procedure, has however, turned out to be too complex for the linguists to handle by themselves. You are therefore assigned the task of writing a program that converts a set of strings to their corresponding Extra Terrestial number sequences.
輸入
On the first line of the input there is a single positive integer N, telling the number of test cases to follow. Each case consists of one line containing the value of p to use during the transcription of the string, followed by the actual string to be transcribed. The only allowed characters in the string are the lower case letters 'a'..'z' and '*' (asterisk). No string will be longer than 70 characters.
輸出
For each transcribed string, output a line with the corresponding list of integers, separated by space, with each integer given in the order of ascending values of i.
樣例輸入
3
31 aaa
37 abc
29 hello*earth
樣例輸出
1 0 0
0 1 0
8 13 9 13 4 27 18 10 12 24 15
高斯消元，求f (k) = ∑0<=i<=n-1aiki (mod p)。
輸入字串str有幾個字母代表有幾個未知數，*代表0，a-z代表1-26，用c[i]表示第i位代表的數
讀入p（模數且為質數），s（下標從0開始），s長度為n
那麼求方程組
的一組合法解
這道題不必判斷無解變元（註釋部分），題目一定是唯一解。
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<math.h>
using namespace std;
#define ll long long
const int MAXN=500;
int a[MAXN][MAXN];//增廣矩陣
int x[MAXN];//解集
int gcd(int a,int b)
{
int t;
while(b!=0)
{
t=b;
b=a%b;
a=t;
}
return a;
}
int lcm(int a,int b)
{
return a/gcd(a,b)*b;
}
int Gauss(int equ,int var,int mod)
{
int i,j,k,max_r;// 當前這列絕對值最大的行.
int col;//當前處理的列
int temp,ta,tb,LCM;
col=0;
for(k=0;k<equ&&col<var;k++,col++)
{
max_r=k;
for(i=k+1;i<equ;i++)
{
if(abs(a[i][col])>abs(a[max_r][col]))
max_r=i;
}
if(max_r!=k)
{
for(j=k;j<var+1;j++)
swap(a[k][j],a[max_r][j]);
}
if(a[k][col]==0)
{
k--;
continue;
}
for(i=k+1;i<equ;i++)
{
// 列舉要刪去的行.
if(a[i][col]!=0)
{
LCM=lcm(abs(a[i][col]),abs(a[k][col]));
ta=LCM/abs(a[i][col]);
tb=LCM/abs(a[k][col]);
if(a[i][col]*a[k][col]<0)
tb=-tb; //異號的情況是相加
for(j=col;j<var+1;j++)
{
a[i][j]=((a[i][j]*ta-a[k][j]*tb)%mod+mod)%mod;
}
}
}
}
/*for (i=k;i<equ;i++)
{
if(a[i][col]!=0) return -1;
}
if(k<var)
{
return var-k;
}*/
for(i=var-1;i>=0;i--)
{
temp=a[i][var];
for(j=i+1;j<var;j++)
{
temp=((temp-a[i][j]*x[j])%mod+mod)%mod;
}
while(temp%a[i][i])
temp+=mod;
temp/=a[i][i];
temp%=mod;
x[i]=temp;
}
return 0;
}
int main()
{
int i,j,equ,var,t,mod,n;
char str[105];
scanf("%d",&t);
while(t--)
{
scanf("%d%s",&mod,str);
n=strlen(str);
equ=var=n;
memset(a,0,sizeof(a));
memset(x,0,sizeof(x));
for(i=0;i<n;i++)
{
if(str[i]=='*')
a[i][n]=0;
else
a[i][n]=str[i]-'a'+1;
a[i][0]=1;
for(j=1;j<n;j++)
{
a[i][j]=(a[i][j-1]*(i+1))%mod;
}
}
int free_num=Gauss(equ,var,mod);
for(i=0;i<var-1;i++)
{
printf("%d ",x[i]);
}
printf("%d\n",x[var-1]);
}
return 0;
}