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  • poj 2065 还是gauss消元

      1 #include <cstdio>
      2 #include <cstring>
      3 #include <iostream>
      4 #include <queue>
      5 #include <vector>
      6 #include <stack>
      7 #include <cmath>
      8 
      9 using namespace std;
     10 
     11 const int M = 10005;
     12 const int maxn = 1000000;
     13 typedef long long LL;
     14 
     15 int a[75][75];
     16 int x[75];
     17 char s[75];
     18 int equ,var,p;
     19 
     20 int gcd(int a,int b)
     21 {
     22     if(b == 0) return a;
     23     else return gcd(b,a%b);
     24 
     25 }
     26 
     27 int lcm(int a,int b)
     28 {
     29     return (a/gcd(a,b))*b;
     30 }
     31 
     32 void gauss()
     33 {
     34     int k,col = 0;
     35     for(k=0;k<equ&&col<var;k++,col++){
     36         int mx_r = k; //找最大的行
     37         for(int i=k+1;i<equ;i++){
     38             if(abs(a[i][col])>abs(a[mx_r][col])) mx_r = i;
     39         }
     40         if(mx_r != k){
     41             for(int i=k;i<=var;i++) swap(a[k][i],a[mx_r][i]);
     42         }
     43         if(a[k][col] == 0){ k--;continue;}
     44         for(int i=k+1;i<equ;i++){
     45             if(a[i][col] != 0){
     46                 int lc = lcm(abs(a[i][col]),abs(a[k][col]));
     47                 int ta = lc/abs(a[i][col]);
     48                 int tb = lc/abs(a[k][col]);
     49                 if(a[i][col]*a[k][col]<0) tb = -tb;
     50                 for(int j=col;j<=var;j++){
     51                    a[i][j] = ((a[i][j]*ta-a[k][j]*tb)%p+p)%p;
     52                 }
     53             }
     54         }
     55     }
     56     int temp;
     57     for(int i=var-1;i>=0;i--){
     58          temp = a[i][var];
     59         for(int j=i+1;j<var;j++){
     60             if(a[i][j] != 0) temp -= a[i][j]*x[j];
     61             temp = (temp%p+p)%p;
     62         }
     63         while(temp % a[i][i]!=0) temp+=p;
     64         x[i] = (temp/a[i][i])%p;
     65     }
     66     for(int i=0;i<var-1;i++){
     67         printf("%d ",x[i]);
     68     }
     69     printf("%d
    ",x[var-1]);
     70 }
     71 
     72 int pow_mod(int a,int n,int MOD)
     73 {
     74     int ret=1;
     75     int temp=a%MOD;
     76     while(n)
     77     {
     78         if(n&1){ret*=temp;ret%=MOD;}
     79         temp*=temp;
     80         temp%=MOD;
     81         n>>=1;
     82     }
     83     return ret;
     84 }
     85 
     86 void init()
     87 {
     88 
     89     memset(x,0,sizeof(x));
     90     memset(a,0,sizeof(a));
     91     memset(s,0,sizeof(s));
     92     scanf("%d",&p);
     93     scanf("%s",s);
     94     int len = strlen(s);
     95     equ = var = len;
     96     for(int i=0;i<len;i++){
     97         if(s[i] == '*') a[i][var] = 0;
     98         else a[i][var] = s[i]-'a'+1;
     99         for(int j=0;j<len;j++){
    100             a[i][j] = pow_mod(i+1,j,p);
    101         }
    102     }
    103 }
    104 
    105 
    106 int main()
    107 {
    108     int t;
    109     scanf("%d",&t);
    110     while(t--){
    111         init();
    112         gauss();
    113     }
    114 
    115     return 0;
    116 }
    View Code
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  • 原文地址:https://www.cnblogs.com/lmlyzxiao/p/4959214.html
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