可能是目前做的这类题都比较简单，所以几乎都是套上模板就差不多了.....
这里copy一份czyuan大神的模板
1 /* 用于求整数解得方程组. */ 2 #include <iostream> 3 #include <string> 4 #include <cmath> 5 using namespace std; 6 const int maxn = 105; 7 int equ, var; // 有equ个方程，var个变元。增广阵行数为equ, 分别为0到equ - 1，列数为var + 1，分别为0到var. 8 int a[maxn][maxn]; 9 int x[maxn]; // 解集. 10 bool free_x[maxn]; // 判断是否是不确定的变元. 11 int free_num; 12 void Debug(void) 13 { 14 int i, j; 15 for (i = 0; i < equ; i++) 16 { 17 for (j = 0; j < var + 1; j++) 18 { 19 cout << a[i][j] << " "; 20 } 21 cout << endl; 22 } 23 cout << endl; 24 } 25 inline int gcd(int a, int b) 26 { 27 int t; 28 while (b != 0) 29 { 30 t = b; 31 b = a % b; 32 a = t; 33 } 34 return a; 35 } 36 inline int lcm(int a, int b) 37 { 38 return a * b / gcd(a, b); 39 } 40 // 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，并返回自由变元的个数) 41 int Gauss(void) 42 { 43 int i, j, k; 44 int max_r; // 当前这列绝对值最大的行. 45 int col; // 当前处理的列. 46 int ta, tb; 47 int LCM; 48 int temp; 49 int free_x_num; 50 int free_index; 51 // 转换为阶梯阵. 52 col = 0; // 当前处理的列. 53 for (k = 0; k < equ && col < var; k++, col++) 54 { // 枚举当前处理的行. 55 // 找到该col列元素绝对值最大的那行与第k行交换.(为了在除法时减小误差) 56 max_r = k; 57 for (i = k + 1; i < equ; i++) 58 { 59 if (abs(a[i][col]) > abs(a[max_r][col])) max_r = i; 60 } 61 if (max_r != k) 62 { // 与第k行交换. 63 for (j = k; j < var + 1; j++) swap(a[k][j], a[max_r][j]); 64 } 65 if (a[k][col] == 0) 66 { // 说明该col列第k行以下全是0了，则处理当前行的下一列. 67 k--; continue; 68 } 69 for (i = k + 1; i < equ; i++) 70 { // 枚举要删去的行. 71 if (a[i][col] != 0) 72 { 73 LCM = lcm(abs(a[i][col]), abs(a[k][col])); 74 ta = LCM / abs(a[i][col]), tb = LCM / abs(a[k][col]); 75 if (a[i][col] * a[k][col] < 0) tb = -tb; // 异号的情况是两个数相加. 76 for (j = col; j < var + 1; j++) 77 { 78 a[i][j] = a[i][j] * ta - a[k][j] * tb; 79 } 80 } 81 } 82 } 83 Debug(); 84 // 1. 无解的情况: 化简的增广阵中存在(0, 0, ..., a)这样的行(a != 0). 85 for (i = k; i < equ; i++) 86 { // 对于无穷解来说，如果要判断哪些是自由变元，那么初等行变换中的交换就会影响，则要记录交换. 87 if (a[i][col] != 0) return -1; 88 } 89 // 2. 无穷解的情况: 在var * (var + 1)的增广阵中出现(0, 0, ..., 0)这样的行，即说明没有形成严格的上三角阵. 90 // 且出现的行数即为自由变元的个数. 91 if (k < var) 92 { 93 // 首先，自由变元有var - k个，即不确定的变元至少有var - k个. 94 for (i = k - 1; i >= 0; i--) 95 { 96 // 第i行一定不会是(0, 0, ..., 0)的情况，因为这样的行是在第k行到第equ行. 97 // 同样，第i行一定不会是(0, 0, ..., a), a != 0的情况，这样的无解的. 98 free_x_num = 0; // 用于判断该行中的不确定的变元的个数，如果超过1个，则无法求解，它们仍然为不确定的变元. 99 for (j = 0; j < var; j++) 100 { 101 if (a[i][j] != 0 && free_x[j]) free_x_num++, free_index = j; 102 } 103 if (free_x_num > 1) continue; // 无法求解出确定的变元. 104 // 说明就只有一个不确定的变元free_index，那么可以求解出该变元，且该变元是确定的. 105 temp = a[i][var]; 106 for (j = 0; j < var; j++) 107 { 108 if (a[i][j] != 0 && j != free_index) temp -= a[i][j] * x[j]; 109 } 110 x[free_index] = temp / a[i][free_index]; // 求出该变元. 111 free_x[free_index] = 0; // 该变元是确定的. 112 } 113 return var - k; // 自由变元有var - k个. 114 } 115 // 3. 唯一解的情况: 在var * (var + 1)的增广阵中形成严格的上三角阵. 116 // 计算出Xn-1, Xn-2 ... X0. 117 for (i = var - 1; i >= 0; i--) 118 { 119 temp = a[i][var]; 120 for (j = i + 1; j < var; j++) 121 { 122 if (a[i][j] != 0) temp -= a[i][j] * x[j]; 123 } 124 if (temp % a[i][i] != 0) return -2; // 说明有浮点数解，但无整数解. 125 x[i] = temp / a[i][i]; 126 } 127 return 0; 128 } 129 int main(void) 130 { 131 //freopen("Input.txt", "r", stdin); 132 int i, j; 133 while (scanf("%d %d", &equ, &var) != EOF) 134 { 135 memset(a, 0, sizeof(a)); 136 memset(x, 0, sizeof(x)); 137 memset(free_x, 1, sizeof(free_x)); // 一开始全是不确定的变元. 138 for (i = 0; i < equ; i++) 139 for (j = 0; j < var + 1; j++) 140 scanf("%d", &a[i][j]); 141 // Debug(); 142 free_num = Gauss(); 143 if (free_num == -1) printf("无解! "); 144 else if (free_num == -2) printf("有浮点数解，无整数解! "); 145 else if (free_num > 0) 146 { 147 printf("无穷多解! 自由变元个数为%d ", free_num); 148 for (i = 0; i < var; i++) 149 { 150 if (free_x[i]) printf("x%d 是不确定的 ", i + 1); 151 else printf("x%d: %d ", i + 1, x[i]); 152 } 153 } 154 else 155 { 156 for (i = 0; i < var; i++) 157 { 158 printf("x%d: %d ", i + 1, x[i]); 159 } 160 } 161 printf(" "); 162 } 163 return 0; 164 } 165 166 /* czyuan原创，转载请注明出处。*/
不过......我一般都是将这个精简过的代码改一下用（里面要用到求lcm最小公倍数和gcd最大共约数的函数）
1 // 高斯消元法解方程组(Gauss-Jordan elimination). 2 //未测试 3 //equ表示方程数，var表示变量数 4 //-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 5 6 int a[N+5][N+5], x[N+5]; //a存储系数矩阵，x存储解 7 8 int gauss(int equ, int var) 9 { 10 int row = 0, col = 0; 11 while (row < equ && col < var){ 12 int flag = row; 13 for (int i = row+1; i < equ; ++ i) 14 if (abs(a[i][col]) > abs(a[flag][col])) 15 flag = i; 16 if (flag != row) for (int i = col; i <= var; ++ i) 17 swap (a[row][i], a[flag][i]); 18 if (!a[row][col]){ ++ col; continue;} 19 20 for (int i = row+1; i < equ; ++ i){ 21 if (a[i][col] == 0) continue; 22 23 int tmp = lcm(abs(a[i][col]), abs(a[row][col])); 24 int ta = tmp / abs(a[i][col]); 25 int tb = tmp / abs(a[row][col]); 26 if (a[i][col] * a[row][col] < 0) tb = -tb; 27 for (int j = col; j <= var; ++ j){ 28 a[i][j] = (a[i][j] * ta - a[row][j]*tb) % 7; 29 if (a[i][j] < 0) a[i][j] += 7; 30 } 31 } 32 ++ row; ++ col; 33 } 34 35 for (int i = row; i < equ; ++ i) 36 if (a[i][var]) return -1; 37 if (row < var) return var - row; 38 for (int i = var-1; i >= 0; -- i){ 39 int tmp = a[i][var]; 40 for (int j = i+1; j < var; ++ j) 41 tmp -= a[i][j]*x[j]; 42 if (tmp % a[i][i]) return -2; 43 x[i] = tmp / a[i][i]; 44 } 45 return 0; 46 }
1、UVa 10828 Back to Kernighan-Ritchie
2、POJ 2947 Widget Factory
1 /* 2 * Author: Plumrain 3 * Created Time: 2013-10-07 16:41 4 * File Name: math-POJ-2947.cpp 5 */ 6 #include<iostream> 7 #include<cstdio> 8 #include<cstring> 9 #include<cmath> 10 #include<algorithm> 11 12 using namespace std; 13 14 #define CLR(x) memset(x, 0, sizeof(x)) 15 16 const int N = 300; 17 18 int a[N+5][N+5], x[N+5]; 19 char ann[7] = {'O', 'U', 'E', 'H', 'R', 'A', 'U'}; 20 21 inline int gcd(int a, int b){ 22 return b == 0 ? a : gcd(b, a%b); 23 } 24 25 inline int lcm(int a, int b){ 26 return a * b / gcd(a, b); 27 } 28 29 int change(char a, char b) 30 { 31 for (int i = 0; i < 7; ++ i) 32 if (i != 1 && i != 6 && ann[i] == b) return i; 33 if (a == 'T') return 1; 34 return 6; 35 } 36 37 void init(int equ, int var) 38 { 39 CLR (a); 40 char s1[10], s2[10]; 41 int k; 42 for (int i = 0; i < equ; ++ i){ 43 scanf ("%d%s%s", &k, s1, s2); 44 a[i][var] = (change(s2[0], s2[1]) - change(s1[0], s1[1]) + 1) % 7; 45 if (a[i][var] < 0) a[i][var] += 7; 46 int tmp; 47 while (k--){ 48 scanf ("%d", &tmp); 49 ++ a[i][tmp-1]; 50 if (a[i][tmp-1] == 7) a[i][tmp-1] = 0; 51 } 52 } 53 } 54 55 int gauss(int equ, int var) 56 { 57 int row = 0, col = 0; 58 while (row < equ && col < var){ 59 int flag = row; 60 for (int i = row+1; i < equ; ++ i) 61 if (abs(a[i][col]) > abs(a[flag][col])) 62 flag = i; 63 if (flag != row) for (int i = col; i <= var; ++ i) 64 swap (a[row][i], a[flag][i]); 65 if (!a[row][col]){ ++ col; continue;} 66 67 for (int i = row+1; i < equ; ++ i){ 68 if (a[i][col] == 0) continue; 69 70 int tmp = lcm(abs(a[i][col]), abs(a[row][col])); 71 int ta = tmp / abs(a[i][col]); 72 int tb = tmp / abs(a[row][col]); 73 if (a[i][col] * a[row][col] < 0) tb = -tb; 74 for (int j = col; j <= var; ++ j){ 75 a[i][j] = (a[i][j] * ta - a[row][j]*tb) % 7; 76 if (a[i][j] < 0) a[i][j] += 7; 77 } 78 } 79 ++ row; ++ col; 80 } 81 82 for (int i = row; i < equ; ++ i) 83 if (a[i][var]) return -1; 84 if (row < var) return var - row; 85 for (int i = var-1; i >= 0; -- i){ 86 int tmp = a[i][var]; 87 for (int j = i+1; j < var; ++ j) 88 tmp = ((tmp - a[i][j]*x[j])%7 + 7) % 7; 89 while (tmp % a[i][i] != 0) tmp += 7; 90 x[i] = (tmp / a[i][i]) % 7; 91 } 92 return 0; 93 } 94 95 int main() 96 { 97 int equ, var; 98 while (scanf ("%d%d", &var, &equ) != EOF && equ){ 99 init(equ, var); 100 101 int ans = gauss(equ, var); 102 if (!ans){ 103 for (int i = 0; i < var; ++ i) 104 if (x[i] <= 2) x[i] += 7; 105 for (int i = 0; i < var-1; ++ i) 106 printf ("%d ", x[i]); 107 printf ("%d ", x[var-1]); 108 } 109 else if (ans == -1) 110 printf ("Inconsistent data. "); 111 else 112 printf ("Multiple solutions. "); 113 } 114 return 0; 115 }
3、POJ 2065 SETI
1 /* 2 * Author: Plumrain 3 * Created Time: 2013-10-08 13:36 4 * File Name: math-POJ-2065.cpp 5 */ 6 #include<iostream> 7 #include<cstdio> 8 #include<cstring> 9 #include<cmath> 10 #include<algorithm> 11 12 using namespace std; 13 14 #define CLR(x) memset(x, 0, sizeof(x)) 15 16 const int N = 70; 17 18 int a[N+5][N+5], mod, n, x[N+5]; 19 20 int gcd(int a, int b) 21 { 22 return b ? gcd(b, a%b) : a; 23 } 24 25 int lcm(int a, int b) 26 { 27 return a * b / gcd(a, b); 28 } 29 30 void init() 31 { 32 CLR (a); CLR (x); 33 char s[N+5]; CLR (s); 34 scanf ("%d%s", &mod, s); 35 n = strlen(s); 36 for (int i = 0; i < n; ++ i) 37 if (s[i] != '*') a[i][n] = s[i] - 'a' + 1; 38 39 for (int i = 0; i < n; ++ i){ 40 int tmp = 1, tim = 0; 41 while (tim < n){ 42 a[i][tim++] = tmp; 43 tmp = (tmp * (i+1)) % mod; 44 } 45 } 46 } 47 48 int gauss() 49 { 50 int row = 0, col = 0; 51 while (row < n && col < n){ 52 int flag = row; 53 for (int i = row+1; i < n; ++ i) 54 if (abs(a[i][col]) > abs(a[flag][col])) 55 flag = i; 56 if (flag != row) for (int i = col; i <= n; ++ i) 57 swap(a[flag][i], a[row][i]); 58 if (!a[row][col]) {++ col; continue;} 59 60 for (int i = row+1; i < n; ++ i){ 61 if (!a[i][col]) continue; 62 63 int tmp = lcm(abs(a[i][col]), abs(a[row][col])); 64 int ta = tmp / abs(a[i][col]); 65 int tb = tmp / abs(a[row][col]); 66 if (a[i][col] * a[row][col] < 0) tb = -tb; 67 for (int j = col; j <= n; ++ j) 68 a[i][j] = ((a[i][j] * ta - a[row][j] * tb)%mod + mod) % mod; 69 } 70 ++ col; ++ row; 71 } 72 73 for (int i = n-1; i >= 0; -- i){ 74 int tmp = a[i][n]; 75 for (int j = i+1; j < n; ++ j) 76 tmp = ((tmp - a[i][j] * x[j]) % mod + mod) % mod; 77 if (a[i][i] < 0){a[i][i] = -a[i][i]; tmp = -tmp;} 78 while (tmp % a[i][i]) tmp += mod; 79 x[i] = (tmp / a[i][i]) % mod; 80 if (x[i] < 0) x[i] += mod; 81 } 82 return 0; 83 } 84 85 int main() 86 { 87 int T; 88 scanf ("%d", &T); 89 while (T--){ 90 init(); 91 92 int ans = gauss(); 93 94 for (int i = 0; i < n; ++ i){ 95 if (i) printf (" "); 96 printf ("%d", x[i]); 97 } 98 printf (" "); 99 } 100 return 0; 101 }
以上三道题，感觉都是直接用模板解就好了，也没什么特别需要想的东西。
4、POJ 1166 The Clocks
这道题，感觉应该用搜索做的....看见网上有人说暴力可以过，有人说用高斯消元也行。但是个人感觉高斯消元应该不行啊，毕竟是要求最优解。而且，在用高斯消元的过程中碰到了一个问题，就是将当前要处理的行与别的行交换时，不能交换a[row][col]绝对值最大的行，也不能交换最后一个a[i][col]非零的行，而按照我下面的写法就恰好能过.....实在不明白原因，猜测这道题是卡数据过的- -。
1 /* 2 * Author: Plumrain 3 * Created Time: 2013-10-08 16:26 4 * File Name: math-POJ-1166.cpp 5 */ 6 #include<iostream> 7 #include<cstdio> 8 #include<cstring> 9 10 using namespace std; 11 12 #define CLR(x) memset(x, 0, sizeof(x)) 13 #define out(x) cout<<#x<<":"<<(x)<<endl 14 #define tst(a) cout<<#a<<endl 15 16 int x[10], sum; 17 int a[10][10]={ 18 }; 19 20 void gauss() 21 { 22 sum = 0; 23 int row = 0, col = 0; 24 while (row < 9 && col < 9){ 25 int flag = row; 26 if (!a[row][col]) 27 for (int i = row+1; i < 9; ++ i) 28 if (a[i][col] && !a[flag][col]) flag = i; 29 if (flag != row) for (int i = col; i <= 9; ++ i) 30 swap (a[row][i], a[flag][i]); 31 if (!a[row][col]){++ col; continue;} 32 33 for (int i = row+1; i < 9; ++ i){ 34 int ta = a[i][col], tb = a[row][col]; 35 for (int j = col; j <= 9; ++ j){ 36 a[i][j] = (a[i][j] * tb - a[row][j] * ta) % 4; 37 if (a[i][j] < 0) a[i][j] += 4; 38 } 39 } 40 ++ col; ++ row; 41 } 42 43 for (int i = 8; i >= 0; -- i){ 44 int tmp = a[i][9]; 45 for (int j = i+1; j < 9; ++ j) 46 tmp -= a[i][j] * x[j]; 47 tmp %= 4; if (tmp < 0) tmp += 4; 48 for (x[i] = 0; x[i] < 4; ++ x[i]) 49 if ((x[i]*a[i][i]%4 + 4) % 4 == tmp) 50 break; 51 x[i] %= 4; 52 sum += x[i]; 53 } 54 } 55 56 int main() 57 { 58 CLR (a); 59 a[0][0]=a[1][0]=a[3][0]=a[4][0]=1; 60 a[0][1]=a[1][1]=a[2][1]=1; 61 a[1][2]=a[2][2]=a[4][2]=a[5][2]=1; 62 a[0][3]=a[3][3]=a[6][3]=1; 63 a[1][4]=a[3][4]=a[4][4]=a[5][4]=a[7][4]=1; 64 a[2][5]=a[5][5]=a[8][5]=1; 65 a[3][6]=a[4][6]=a[6][6]=a[7][6]=1; 66 a[6][7]=a[7][7]=a[8][7]=1; 67 a[4][8]=a[5][8]=a[7][8]=a[8][8]=1; 68 69 for (int i = 0; i < 9; ++ i){ 70 scanf ("%d", &a[i][9]); 71 a[i][9] = (4 - a[i][9]) % 4; 72 } 73 74 gauss(); 75 76 for (int i = 0; i < 9; ++ i) 77 while (x[i]){ 78 printf ("%d", i+1); 79 -- x[i]; -- sum; 80 printf (sum>0 ? " " : " "); 81 } 82 return 0; 83 }
5、POJ 1222 EXTENDED LIGHTS OUT
题意：POJ的Discuss里面有人说可以证明此题只有唯一解。我不会证。此题可以用高斯消元的方法做，当然也可以用状态压缩暴力枚举第一行 + 递推的方式做。我比较偷懒，选择了后者。
1 /* 2 * Author: Plumrain 3 * Created Time: 2013-10-08 18:52 4 * File Name: simulate-POJ-1222.cpp 5 */ 6 #include<iostream> 7 #include<cstdio> 8 #include<cstring> 9 #include<vector> 10 #include<utility> 11 12 using namespace std; 13 14 #define CLR(x) memset(x, 0, sizeof(x)) 15 #define PB push_back 16 #define out(x) cout<<#x<<":"<<(x)<<endl 17 #define tst(a) cout<<#a<<endl 18 19 int a[10][10], tmp[10][10], num[10][10]; 20 vector<pair<int, int> > ans; 21 22 void init() 23 { 24 ans.clear(); 25 for (int i = 0; i < 5; ++ i) 26 for (int j = 0; j < 6; ++ j) 27 scanf ("%d", &a[i][j]); 28 29 for (int i = 0; i < 5; ++ i) 30 for (int j = 0; j < 6; ++ j) 31 tmp[i][j] = a[i][j]; 32 } 33 34 void change(int x, int y) 35 { 36 ans.PB (make_pair(x, y)); 37 a[x][y] = 1 - a[x][y]; 38 if (x) a[x-1][y] = 1 - a[x-1][y]; 39 if (x < 4) a[x+1][y] = 1 - a[x+1][y]; 40 if (y) a[x][y-1] = 1 - a[x][y-1]; 41 if (y < 5) a[x][y+1] = 1 - a[x][y+1]; 42 } 43 44 void all_init() 45 { 46 ans.clear(); 47 for (int i = 0; i < 5; ++ i) 48 for (int j = 0; j < 6; ++ j) 49 a[i][j] = tmp[i][j]; 50 } 51 52 bool all_off(int x) 53 { 54 for (int i = 0; i < 6; ++ i) 55 if (a[x][i]) return false; 56 return true; 57 } 58 59 void gao() 60 { 61 for (int i = 0; i < (1<<6); ++ i){ 62 for (int j = 0; j < 6; ++ j) 63 if (i & (1<<j)) change(0, j); 64 65 for (int j = 1; j < 5; ++ j) 66 for (int k = 0; k < 6; ++ k) 67 if (a[j-1][k]) change(j, k); 68 69 if (all_off(4)) return; 70 all_init(); 71 } 72 } 73 74 int main() 75 { 76 int T, test = 0; 77 scanf ("%d", &T); 78 while (T--){ 79 printf ("PUZZLE #%d ", ++ test); 80 81 init(); 82 gao(); 83 84 CLR (num); 85 for (int i = 0; i < ans.size(); ++ i) 86 num[ans[i].first][ans[i].second] = 1; 87 88 for (int i = 0; i < 5; ++ i){ 89 for (int j = 0; j < 5; ++ j) 90 printf ("%d ", num[i][j]); 91 printf ("%d ", num[i][5]); 92 } 93 } 94 return 0; 95 }