PS. 看了大神的题解，发现确实可以用m个未知数的高斯消元做。因为确定了第一行的情况，之后所有行的情况都可以根据第一行推。 这样复杂度直接变成O(m*m*m)
知道了是高斯消元后，其实只要稍加处理，就可以解决带模的情况。
1 是在进行矩阵行变化的时候，取模。
2 最后的除法用逆元。（因为a[i][i]必定非0 且小于模数）
然后对于无穷多解的情况，只需要将那些列全为0的未知数定义一个固定值。（这里设的是0）其余操作不变。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <set> #include <map> #include <string> #define CL(a,num) memset((a),(num),sizeof(a)) #define iabs(x) ((x) > 0 ? (x) : -(x)) #define Min(a,b) (a) > (b)? (b):(a) #define Max(a,b) (a) > (b)? (a):(b) #define ll long long #define inf 0x7f7f7f7f #define MOD 100000007 #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define test puts("<------------------->") #define maxn 100007 #define M 100007 #define N 1010 using namespace std; //freopen("din.txt","r",stdin); int a[N][N],X[N];//分别记录增广矩阵和解集 int free_x[N];//记录自由变量 int equ,var;//分别表示方程组的个数和变量的个数 int GCD(int x,int y){ if (y == 0) return x; return GCD(y,x%y); } int LCM(int x,int y){ return x/GCD(x,y)*y; } void Debug(void) { int i, j; for (i = 0; i < equ; i++) { for (j = 0; j < var + 1; j++) { cout << a[i][j] << " "; } cout << endl; } cout << endl; } // 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， //并返回自由变元的个数) //ax + by = gcd(a,b) //传入固定值a,b.放回 d=gcd(a,b), x , y void extendgcd(ll a,ll b,ll &d,ll &x,ll &y) { if(b==0){d=a;x=1;y=0;return;} extendgcd(b,a%b,d,y,x); y-=x*(a/b); } //Ax=1(mod M)，gcd(A,M)==1 //输入：10^18>=A,M>=1 //输出：返回x的范围是[1,M-1] ll GetNi(ll A,ll MM) { ll rex=0,rey=0; ll td=0; extendgcd(A,MM,td,rex,rey); return (rex%MM+MM)%MM; } int Guass(){ int i,j,k,col; CL(X,0); CL(free_x,1); for (k = 0,col = 0; k < equ && col < var; k++, col++){ int max_r = k; for (i = k + 1; i < equ; ++i){ if (iabs(a[i][col]) > iabs(a[max_r][col])) max_r = i; } if (max_r != k){ for (i = k; i < var + 1; ++i) swap(a[k][i],a[max_r][i]); } if (a[k][col] == 0){ //可以随意定义的变量 X[col] = 0;//强制赋值为0 free_x[col] = 0; k--; //cout<<k<<endl; continue; } for (i = k + 1; i < equ; ++i){ if (a[i][col] != 0){ int lcm = LCM(a[k][col],a[i][col]); int ta = lcm/iabs(a[i][col]); int tb = lcm/iabs(a[k][col]); if (a[i][col]*a[k][col] < 0) tb = -tb; for (j = col; j < var + 1; ++j){ a[i][j] = ((ta*a[i][j] - tb*a[k][j])%3+3)%3; } } } } //Debug(); // 1. 无解的情况: for (i = k; i < equ; ++i){ if (a[i][col] != 0) return -1; } // 2. 无穷解的情况: if (k < var){ int num = 0,freeidx=0; for (i = k - 1; i >= 0; --i){ num = 0; int tmp = a[i][var]; for (j = 0; j < var; ++j){ if (a[i][j] != 0 && free_x[j]){ num++; freeidx = j; } } if (num > 1) continue; tmp = a[i][var]; for (j = 0; j < var; ++j){ if (a[i][j] && j != freeidx) tmp -= a[i][j]*X[j]; } //这里也要 int k2 = (tmp%3+3)%3; int k1 = (a[i][freeidx]%3+3)%3; if(k1!=0) { X[freeidx] = k2*(int)GetNi(k1, 3); } else { X[freeidx] = 0; //printf("X[%d]为任意？ ",i); } //X[freeidx] = tmp/a[i][freeidx]; free_x[freeidx] = 0; } return var - k; } // 3. 唯一解的情况: for (i = k - 1; i >= 0; --i){ int tmp = a[i][var]; for (j = i + 1; j < var; ++j){ tmp -= a[i][j]*X[j]; } //强行搞一发 int k2 = (tmp%3+3)%3; int k1 = (a[i][i]%3+3)%3; if(k1!=0) { X[i] = k2*(int)GetNi(k1, 3); } else { X[i] = 0; //printf("X[%d]为任意？ ",i); } //X[i] = tmp/a[i][i];//不整除？ } return 0; } int g[33][33]; int getid(int i,int j,int n,int m) { return (i-1)*m + (j-1); } int up[4]={1,-1,0,0}; int rl[4]={0,0,1,-1}; int main(){ //freopen("din.txt","r",stdin); int i,j; int T; cin>>T; while(T--) { int n,m; scanf("%d%d",&n,&m); for(i=1;i<=n;i++) for(j=1;j<=m;j++) scanf("%d",&g[i][j]); equ = n*m; var = n*m; memset(a,0,sizeof(a)); int id = 0; for(i=1;i<=n;i++) for(j=1;j<=m;j++) { for(int k=0;k<4;k++) { int ti = i+up[k]; int tj = j+rl[k]; if( ti>=1&&ti<=n && tj>=1&&tj<=m ) { int tid = getid(ti, tj, n, m); a[id][tid] = 1; } } a[id][id] = 2; a[id][n*m] = (3-g[i][j])%3; id++; } CL(X,0); CL(free_x,0); //for (i = 0; i < equ; ++i) //for (j = 0; j < var + 1; ++j) scanf("%d",&a[i][j]); //Debug(); int free_num = Guass(); if (free_num == -1) printf("无解！ "); else if (free_num == -2) printf("无整数解 "); else { // else if (free_num > 0){ //printf("无穷多解! 自由变元个数为%d ", free_num); //我懂了，我需要确定free_num个数 // for (i = 0; i < var; ++i){ // if (free_x[i]) printf("X%d 是不确定的 ",i + 1); // else printf("X%d %d ",i + 1,X[i]); // } // } // else{ //我觉得可以！ int ans = 0; for (i = 0 ; i < var; ++i){ if(X[i]%3 != 0) ans += (X[i]%3+3)%3; //printf("X%d %d ",i + 1,X[i]); } printf("%d ",ans); for(i=0;i<var;i++) { int tmp = (X[i]%3+3)%3; for(j=0;j<tmp;j++) { int x,y; x = i/m; y = i%m; x++; y++; printf("%d %d ",x,y); } } } //printf(" "); } return 0; } /* 10 3 3 0 0 0 0 0 0 0 0 1 1 2 0 0 2 30 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 */
2. m个未知数的情况
// // main.cpp // hdu5755.1 // // Created by New_Life on 16/8/4. // Copyright © 2016年 chenhuan001. All rights reserved. // #include <iostream> #include <string.h> #include <stdio.h> #include <algorithm> #include <queue> #include <stack> #include <set> #include <map> #include <string> #define CL(a,num) memset((a),(num),sizeof(a)) #define iabs(x) ((x) > 0 ? (x) : -(x)) #define Min(a,b) (a) > (b)? (b):(a) #define Max(a,b) (a) > (b)? (a):(b) #define ll long long #define inf 0x7f7f7f7f #define MOD 100000007 #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define test puts("<------------------->") #define maxn 100007 #define M 100007 #define N 33 using namespace std; //freopen("din.txt","r",stdin); int a[N][N],X[N];//分别记录增广矩阵和解集 int free_x[N];//记录自由变量 int equ,var;//分别表示方程组的个数和变量的个数 int GCD(int x,int y){ if (y == 0) return x; return GCD(y,x%y); } int LCM(int x,int y){ return x/GCD(x,y)*y; } void Debug(void) { int i, j; for (i = 0; i < equ; i++) { for (j = 0; j < var + 1; j++) { cout << a[i][j] << " "; } cout << endl; } cout << endl; } // 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， //并返回自由变元的个数) //ax + by = gcd(a,b) //传入固定值a,b.放回 d=gcd(a,b), x , y void extendgcd(ll a,ll b,ll &d,ll &x,ll &y) { if(b==0){d=a;x=1;y=0;return;} extendgcd(b,a%b,d,y,x); y-=x*(a/b); } //Ax=1(mod M)，gcd(A,M)==1 //输入：10^18>=A,M>=1 //输出：返回x的范围是[1,M-1] ll GetNi(ll A,ll MM) { ll rex=0,rey=0; ll td=0; extendgcd(A,MM,td,rex,rey); return (rex%MM+MM)%MM; } int Guass(){ int i,j,k,col; CL(X,0); CL(free_x,1); for (k = 0,col = 0; k < equ && col < var; k++, col++){ int max_r = k; for (i = k + 1; i < equ; ++i){ if (iabs(a[i][col]) > iabs(a[max_r][col])) max_r = i; } if (max_r != k){ for (i = k; i < var + 1; ++i) swap(a[k][i],a[max_r][i]); } if (a[k][col] == 0){ //可以随意定义的变量 X[col] = 0;//强制赋值为0 free_x[col] = 0; k--; //cout<<k<<endl; continue; } for (i = k + 1; i < equ; ++i){ if (a[i][col] != 0){ int lcm = LCM(a[k][col],a[i][col]); int ta = lcm/iabs(a[i][col]); int tb = lcm/iabs(a[k][col]); if (a[i][col]*a[k][col] < 0) tb = -tb; for (j = col; j < var + 1; ++j){ a[i][j] = ((ta*a[i][j] - tb*a[k][j])%3+3)%3; } } } } //Debug(); // 1. 无解的情况: for (i = k; i < equ; ++i){ if (a[i][col] != 0) return -1; } // 2. 无穷解的情况: if (k < var){ int num = 0,freeidx=0; for (i = k - 1; i >= 0; --i){ num = 0; int tmp = a[i][var]; for (j = 0; j < var; ++j){ if (a[i][j] != 0 && free_x[j]){ num++; freeidx = j; } } if (num > 1) continue; tmp = a[i][var]; for (j = 0; j < var; ++j){ if (a[i][j] && j != freeidx) tmp -= a[i][j]*X[j]; } //这里也要 int k2 = (tmp%3+3)%3; int k1 = (a[i][freeidx]%3+3)%3; if(k1!=0) { X[freeidx] = k2*(int)GetNi(k1, 3); } else { X[freeidx] = 0; //printf("X[%d]为任意？ ",i); } //X[freeidx] = tmp/a[i][freeidx]; free_x[freeidx] = 0; } return var - k; } // 3. 唯一解的情况: for (i = k - 1; i >= 0; --i){ int tmp = a[i][var]; for (j = i + 1; j < var; ++j){ tmp -= a[i][j]*X[j]; } //强行搞一发 int k2 = (tmp%3+3)%3; int k1 = (a[i][i]%3+3)%3; if(k1!=0) { X[i] = k2*(int)GetNi(k1, 3); } else { X[i] = 0; //printf("X[%d]为任意？ ",i); } //X[i] = tmp/a[i][i];//不整除？ } return 0; } int g[33][33]; int save[33][33][33]; int getni(int x) { if(x==1) return 2; if(x==2) return 1; return 0; } void func(int x,int y) { g[x][y] = (g[x][y]+2)%3; g[x+1][y] = (g[x+1][y]+1)%3; g[x][y+1] = (g[x][y+1]+1)%3; g[x-1][y] = (g[x-1][y]+1)%3; g[x][y-1] = (g[x][y-1]+1)%3; } int main() { int T; cin>>T; while(T--) { int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) scanf("%d",&g[i][j]); memset(save,0,sizeof(save)); for(int i=1;i<=m;i++) { save[1][i][i] = 1; } for(int i=2;i<=n;i++) { for(int j=1;j<=m;j++) { for(int k=0;k<=m;k++) { int tmp = (save[i-1][j][k]*2+save[i-1][j+1][k]+save[i-1][j-1][k]) + save[i-2][j][k]; if(k == 0) tmp += g[i-1][j]; tmp %= 3; save[i][j][k] = getni(tmp); } } } //然后构建矩阵 memset(a,0,sizeof(a)); equ = m; var = m; for(int j=1;j<=m;j++) { for(int k=0;k<=m;k++) { int tmp = save[n][j][k]*2+save[n][j+1][k]+save[n][j-1][k] + save[n-1][j][k]; if(k==0) { tmp += g[n][j]; a[j-1][m] = getni(tmp%3); } else a[j-1][k-1] = tmp%3; } } CL(X,0); CL(free_x,0); int free_num = Guass(); if(free_num == -1 || free_num == -2){ cout<<free_num<<" 错误"<<endl; continue;} int ans = 0; int saveans[33][33]; memset(saveans,0,sizeof(saveans)); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) { int tmp = 0; for(int k=0;k<=m;k++) { if(k==0) tmp += save[i][j][0]; else tmp += save[i][j][k]*X[k-1]; tmp %= 3; } ans += tmp; saveans[i][j] = tmp; } printf("%d ",ans); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) { for(int k=0;k<saveans[i][j];k++) { printf("%d %d ",i,j); //func(i,j); } } // for(int i=1;i<=n;i++) // { // for(int j=1;j<=m;j++) printf("%d ",g[i][j]); // printf(" "); // } } return 0; } /* 10 1 3 0 0 1 3 3 0 0 0 0 0 0 0 0 1 1 2 0 0 2 30 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 */