POJ 2947 Widget Factory 环上的Gauss

题意：饰品工厂生产一件饰品需要3-9天，现在知道一些工人的工作时间范围，以及他们生产出来的饰品种类。求每种饰品的生产所需的时间。

题解：

#include<cstdio>#include<iostream>#include<cmath>#include<cstring>#include<algorithm>using namespace std;#define MAXN 310int a[MAXN][MAXN], x[MAXN];int equ, var;char table[10][4] = { "MON", "TUE", "WED", "THU", "FRI", "SAT","SUN" };inline int gcd ( int a, int b ){if ( b == 0 ) return a;return gcd ( b, a % b );}inline int lcm ( int a, int b ){return a * b / gcd(a, b);}void Debug (){    int i, j;    printf("Debug**************************\n");    for ( i = 0; i < equ; i++ )    {        for ( j = 0; j <= var; j++ )            printf("%d ",a[i][j]);        printf("\n");    }}// 高斯消元法解方程组(Gauss-Jordan elimination).//-2表示有浮点数解，但无整数解//-1表示无解//0表示唯一解//大于0表示无穷解，并返回自由变量的个数int Gauss (){int i, j, mr, row, col;int ta, tb, l, tmp;row = col = 0;while ( row < equ && col < var ){mr = row;for ( i = row + 1; i < equ; i++ )if ( abs(a[i][col]) > abs(a[mr][col]) ) mr = i;if ( mr != row )for ( j = col; j <= var; j++ )swap ( a[row][j], a[mr][j] );if ( a[row][col] == 0 ) { col++; continue; }for ( i = row + 1; i < equ; i++ ){if ( a[i][col] != 0 ){l = lcm(abs(a[i][col]), abs(a[row][col]));ta = l / abs(a[i][col]), tb = l / abs(a[row][col]);if ( a[i][col] * a[row][col] < 0 ) tb = -tb;for ( j = col; j <= var; j++ )a[i][j] = ((a[i][j]*ta-a[row][j]*tb)%7+7) % 7;}}row++; col++;}    //Debug();for ( i = row; i < equ; i++ )if ( a[i][var] != 0 ) return -1;if ( row < var )return var - row;for ( i = var - 1; i >= 0; i-- ){tmp = a[i][var];for ( j = i + 1; j < var; j++ )tmp = ((tmp-a[i][j]*x[j])%7+7)%7;        while ( tmp % a[i][i] != 0 ) tmp += 7;x[i] = ( tmp / a[i][i] ) % 7;}return 0;}int getTime ( char st[], char ed[] ){    int s = 0, e  = 0;    for ( int i = 0; i < 7; i++ )    {        if ( strcmp(table[i],st) == 0 ) s = i;        if ( strcmp(table[i],ed) == 0 ) e = i;    }    return ((e - s + 1) % 7 + 7) % 7;}int main(){    char st[10], ed[10];    while ( scanf("%d%d",&var,&equ) && (var||equ) )    {        int i, res, k, kind;        memset(a,0,sizeof(a));        for ( i = 0; i < equ; i++ )        {            scanf("%d %s %s",&k,st,ed);            a[i][var] = getTime (st,ed);            a[i][var] %= 7;            while ( k-- )            {                scanf("%d",&kind);                a[i][kind-1]++;                a[i][kind-1] %= 7;            }        }        //Debug();        res = Gauss();        if ( res == 0 )        {            for ( i = 0; i < var; i++ )                if ( x[i] <= 2 ) x[i] += 7;            for ( i = 0; i < var - 1; i++ )                printf("%d ",x[i]);            printf("%d\n",x[i]);        }        else if ( res == -1 )            printf("Inconsistent data.\n");        else            printf("Multiple solutions.\n");    }    return 0;}