(Template) poj2947 (Gaussian elimination method for solving the congruence equations)

Topic link: https: //vjudge.net/problem/POJ-2947

Meaning of the questions: After the conversion meaning of the questions is known of m congruence equations, find n variables.

Ideas:

　　It is worth learning to use this template lcm elimination of that piece. Note that the answer to the title output between [3,9].

AC Code:

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdlib>
using namespace std;

const int maxn=305;
int n,m,a[maxn][maxn],x[maxn];
char s1[10],s2[10];

int gcd(int a,int b){
    return b?gcd(b,a%b):a;
}

int lcm(int a,int b){
    return a/gcd(a,b)*b;  //In addition to the first post-multiplication 
} 

// Gaussian elimination method for solving equations (GAUSS-Jordan elimination). (
 @ -1 indicates no solution, 0 represents a unique solution, greater than 0 indicates infinite solutions, and returns the number of arguments free)
 // there equ equations, var an argument. Augmented number of rows of the matrix of equ, respectively, 0 to equ-1, the number of columns of var + 1, respectively 0 to var. 
Int GAUSS ( int equ, int  var ) {
     int K, max_r, COL = 0 , TA, TB, the LCM, TEMP;
     for ( int I = 0 ; I < var ; ++ I) { 
        X [I] = 0 ; 
    } 
    for (K = 0 ; K <COL && EQU < var ; K ++, ++ COL) { 
        max_r =k;
         // find the largest absolute value coefficient k-th row and the row exchange 
        for ( int I = k + . 1 ; I <EQU; ++ I) {
             IF (ABS (A [I] [COL])> ABS (A [max_r] [COL])) 
                max_r = I; 
        } 
        IF ! (max_r = K) {
             for ( int I = COL; I < var + . 1 ; ++ I) 
                the swap (A [max_r] [I], A [ K] [I]); 
        } 
        IF (! A [K] [COL]) {
             - K;
             Continue ; 
        } 
        for (int i=k+1;i<equ;++i){
            if(!a[i][col]) continue;
            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; //异号的情况是相加
            for(int j=col;j<var+1;++j){
                a[i][j]=((a[i][j]*ta-a[k][j]*tb)%7+7)%7;
            } 
        } 
    } 
    // no solution conditions 
    for ( int I = K; I <EQU; ++ I) {
         IF (A [I] [COL]) return - . 1 ; 
    } 
    // infinite solutions case where 
    IF (K < var ) {    
         return  var -k;    // returns the number of arguments consisting 
    }
     // the only solution, augmented matrix formed strictly upper triangular matrix 
    for ( int I = var - . 1 ; I> = 0 ; - - I) { 
        TEMP = A [I] [ var ];
        for(int j=i+1;j<var;++j){
            if(!a[i][j]) continue;
            temp-=a[i][j]*x[j];
            temp=(temp%7+7)%7;
        }
        while(temp%a[i][i]!=0) temp+=7;
        x[i]=(temp/a[i][i])%7;
    }
    return 0;
}

int tran(char *s){
    if(strcmp(s,"MON")==0) return 1;
    else if(strcmp(s,"TUE")==0) return 2;
    else if(strcmp(s,"WED")==0) return 3;
    else if(strcmp(s,"THU")==0) return 4;
    else if(strcmp(s,"FRI")==0) return 5;
    else if(strcmp(s,"SAT")==0) return 6;
    else return 7;
}

int main(){
    while(scanf("%d%d",&n,&m),n||m){
        memset(a,0,sizeof(a));
        for(int i=0;i<m;++i){
            int k;
            scanf("%d%s%s",&k,s1,s2);
            a[i][n]=((tran(s2)-tran(s1)+1)%7+7)%7;
            while(k--){
                int t;
                scanf("%d",&t);
                --t;
                ++a[i][t];
                a[i][t]%=7;
            }
        }
        int ans=Gauss(m,n);
        if(ans==0){
            for(int i=0;i<n;++i)
                if(x[i]<=2) x[i]+=7;
            for(int i=0;i<n-1;++i)
                printf("%d ",x[i]);
            printf("%d\n",x[n-1]);
        }
        else if(ans==-1){
            printf("Inconsistent data.\n");
        }
        else{
            printf("Multiple solutions.\n");
        }
    }
    return 0;
}