给出组件的数量n，给出记录的数量m（n就是变元数量，m是方程数量）。每一个记录代表一个方程，求每个组件的生产天数。

高斯消元即可

#include
#include
#include
#include

const int MOD = 7;
const int MAXN = 310;
int a[MAXN][MAXN], x[MAXN];

void debug(int n, int m)
{
for(int i=0; i<n; i++)
{
for(int j=0; j<m; j++)
printf("%d ", a[i][j]);
printf(" %d\n", a[i][m]);
}
puts("**************************************************************");
}

int gcd(int a,int b) //递归算法
{
return b ? gcd(b, a%b) : a;
}

int lcm(int a, int b)
{
return a*b/gcd(a,b);
}

int Guass(int equ,int var)
{
// debug(equ, var);
int row,col;
row=col=0;
while(row=0; i--)
{
int tmp = a[i][var];
for(int j=i+1; j<var; j++)
tmp = ((tmp - x[j]*a[i][j])%MOD + MOD)%MOD;
while(tmp%a[i][i]) tmp += MOD;
x[i] = tmp/a[i][i]%MOD;
}
return 0;
}

int main()
{
// freopen("in.txt", "r", stdin);
int n,m;
mapmp;
mp["MON"] = 1, mp["TUE"] = 2, mp["WED"] = 3;
mp["THU"] = 4, mp["FRI"] = 5, mp["SAT"] = 6, mp["SUN"] = 7;
while(~scanf("%d%d", &n, &m)) //n个变元，m个方程
{
memset(a, 0, sizeof(a));
if(n == 0 && m == 0) break;
for(int i=0; i<m; i++)
{
int k;
char s1[5], s2[5];
scanf("%d", &k);
scanf("%s%s", s1, s2);
a[i][n] = (mp[s2]-mp[s1]+1+MOD)%MOD;
while(k--)
{
int t;
scanf("%d", &t);
a[i][t-1]++;
a[i][t-1] %= MOD;
}
}
int res = Guass(m, n);
if(res == -1)
puts("Inconsistent data.");
else if(res == 1)
puts("Multiple solutions.");
else
{
for(int i=0; i<n; i++)
if(x[i]<3) x[i] += MOD;
for(int i=0; i<n; i++)
printf("%d%c", x[i], i==n-1?'\n':' ');
}
}
return 0;
}