本文主要是介绍POJ2947 DAZE [Gauss]，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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]
如果有解输出解得个数，如果无解Inconsistent data.无穷多组解Multiple solutions.
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]);elseprintf(" %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;
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