POJ2947 DAZE [Gauss]

题目是要求建立一个方程组：

(mat[1][1]*x[1] + mat[1][2]*x[2] + … + mat[1][n]*x[n])%7 =mat[1][n+1]

(mat[2][1]*x[1] + mat[2][2]*x[2] + … + mat[2][n]*x[n])%7 =mat[2][n+1]

…

…

(mat[m][1]*x[1] + mat[m][2]*x[2] + … + mat[m][n]*x[n])%7 =mat[m][n+1]

扯一句，什么时候无解？

系数矩阵的秩 不等于 增广矩阵的秩 时。反映在程序上就是：

for (i = row; i < N; i++)
{
if (a[i][M] != 0)
{
printf("Inconsistent data.\n");
}
}

什么时候无穷多解？

当增广矩阵的秩小于行列式的行数的时候。 反映在程序上就是：

if (row < M)
{
printf("Multiple solutions.\n");
}

好了，程序如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cstdlib>
using namespace std;
int m,n;//n lines
int M,N;
int a[333][333];
int times[333];
int ans[333];
int MOD=7;
int getday(char* s)
{
	if(strcmp(s,"MON")==0) return 1;
	if(strcmp(s,"TUE")==0) return 2;
	if(strcmp(s,"WED")==0) return 3;
	if(strcmp(s,"THU")==0) return 4;
	if(strcmp(s,"FRI")==0) return 5;
	if(strcmp(s,"SAT")==0) return 6;
	if(strcmp(s,"SUN")==0) return 7;
}

int extend_gcd(int A, int B, int &x, int &y)
{
    if (B == 0)
    {
        x = 1, y = 0;
        return A;
    }
    else
    {
        int r = extend_gcd(B, A%B, x, y);
        int t = x;
        x = y;
        y = t - A / B*y;
        return r;
    }
}

int lcm(int A, int B)
{
    int x = 0, y = 0;
    return A*B / extend_gcd(A, B, x, y);
}
void Guass()
{
    int i, j, row, col;
    for (row = 0, col = 0; row < N && col < M; row++, col++)
           {
        for (i = row; i < N; i++)
            if (a[i][col]) break;
        if (i == N)
                      {
            row--;
            continue;
                       }
        if (i != row)
            for (j = 0; j <= M; j++) swap(a[row][j], a[i][j]);
        for (i = row + 1; i < N; i++)
                     {
            if (a[i][col])
                                {
                int LCM = lcm(a[row][col], a[i][col]);//利用最小公倍数去化上三角
                int ch1 = LCM / a[row][col], ch2 = LCM / a[i][col];
                for (j = col; j <= M; j++)
                    a[i][j] = ((a[i][j] * ch2 - a[row][j] * ch1)%MOD + MOD)%MOD;
                                }
                     }
         }
   for (i = row; i < N; i++)//无解
        {
        if (a[i][M] != 0)
                      {
            printf("Inconsistent data.\n");
            return;
                      }
         }
   if (row < M)//无穷多解
        {
        printf("Multiple solutions.\n");
        return;
        }
        //唯一解时
   for (i = M - 1; i >= 0; i--)
        {
        int ch = 0;
        for (j = i + 1; j < M; j++)
                       {
            ch = (ch + ans[j] * a[i][j] % MOD)%MOD;
                       }
        int last = ((a[i][M] - ch)%MOD + MOD)%MOD;
        int x = 0, y = 0;
        int d = extend_gcd(a[i][i], MOD, x, y);
        x %= MOD;
        if (x < 0) x += MOD;
                ans[i] = last*x / d%MOD;
        if (ans[i] < 3) ans[i] += 7;
          }
   for (int i = 0; i < M; i++)
        {
        if (i == 0)
                 printf("%d", ans[i]);
        else
                 printf(" %d", ans[i]);
        }
   printf("\n");
}

int main()
{
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		if(m==0&&n==0)
			break;
		M=m,N=n;
		memset(a,0,sizeof(a));
		memset(times,0,sizeof(times));
		memset(ans,0,sizeof(ans));
		for(int i=0;i<n;i++)
		{
			int prodnum;char str1[11],str2[11];
			scanf("%d%s%s",&prodnum,str1,str2);
			for(int j=0;j<prodnum;j++)
			{
				int tmp;
				scanf("%d",&tmp);
				a[i][tmp-1]++;
				a[i][tmp-1]%=MOD;
			}
			a[i][m]=(getday(str2)-getday(str1)+1+MOD)%MOD;
		}
		Guass();
	}
	return 0;
}

POJ2947 DAZE [Gauss],布布扣,bubuko.com

POJ2947 DAZE [Gauss]

原文：http://blog.csdn.net/u011775691/article/details/35393507