高斯消元 o(n^3) 取摸和不取摸

    #include<bits/stdc++.h>
    using namespace std;
    const int MAXN=50;
    int a[MAXN][MAXN];//增广矩阵
    int x[MAXN];//解集
    bool free_x[MAXN];//标记是否是不确定的变元
    int gcd(int a,int b){ if(b == 0) return a; else return gcd(b,a%b);}
    inline int lcm(int a,int b){return a/gcd(a,b)*b;}//先除后乘防溢出}
    // 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)
    //有equ个方程，var个变元。增广矩阵行数为equ,分别为0到equ-1,列数为var+1,分别为0到var.
    int Gauss(int equ,int var)
    {
        int i,j,k;
        int max_r;// 当前这列绝对值最大的行.
        int col;//当前处理的列
        int ta,tb;
        int LCM;
        int temp;
        int free_x_num;
        int free_index;

        for(int i=0;i<=var;i++){
            x[i]=0;
            free_x[i]=true;
        }

        //转换为阶梯阵.
        col=0; // 当前处理的列
        for(k = 0;k < equ && col < var;k++,col++){// 枚举当前处理的行.
        // 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)
            max_r=k;
            for(i=k+1;i<equ;i++){
                if(abs(a[i][col])>abs(a[max_r][col])) max_r=i;
            }
            if(max_r!=k){// 与第k行交换.
                for(j=k;j<var+1;j++) swap(a[k][j],a[max_r][j]);
            }
            if(a[k][col]==0){// 说明该col列第k行以下全是0了，则处理当前行的下一列.
                k--;
                continue;
            }
            for(i=k+1;i<equ;i++){// 枚举要删去的行.
                if(a[i][col]!=0){
                    LCM = lcm(abs(a[i][col]),abs(a[k][col]));
                    ta = LCM/abs(a[i][col]);
                    tb = LCM/abs(a[k][col]);
                    if(a[i][col]*a[k][col]<0)tb=-tb;//异号的情况是相加
                    for(j=col;j<var+1;j++){
                        a[i][j] = a[i][j]*ta-a[k][j]*tb;
                    }
                }
            }
        }
        // 1. 无解的情况: 化简的增广阵中存在(0, 0, ..., a)这样的行(a != 0).
        for (i = k; i < equ; i++){ // 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.
            if (a[i][col] != 0) return -1;
        }
        // 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.
        // 且出现的行数即为自由变元的个数.
        if (k < var){
            return var - k; // 自由变元有var - k个.
        }
        // 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.
        // 计算出Xn-1, Xn-2 ... X0.
        for (i = var - 1; i >= 0; i--){
            temp = a[i][var];
            for (j = i + 1; j < var; j++){
                if (a[i][j] != 0) temp -= a[i][j] * x[j];
            }
            if (temp % a[i][i] != 0) return -2; // 说明有浮点数解，但无整数解.
            x[i] = temp / a[i][i];
        }
        return 0;
    }
    int main(void){
    //    freopen("in.txt", "r", stdin);
    //    freopen("out.txt","w",stdout);
        int i, j;
        int equ,var;
        while (scanf("%d %d", &equ, &var) != EOF){
            memset(a, 0, sizeof(a));
            for (i = 0; i < equ; i++){
                for (j = 0; j < var + 1; j++){
                    scanf("%d", &a[i][j]);
                }
            }
            int free_num = Gauss(equ,var);
            if (free_num == -1) printf("无解!
");
            else if (free_num == -2) printf("有浮点数解，无整数解!
");
            else if (free_num > 0){
                printf("无穷多解! 自由变元个数为%d
", free_num);
                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{
                for (i = 0; i < var; i++){
                    printf("x%d: %d
", i + 1, x[i]);
                }
            }
            printf("
");
        }
        return 0;
    }

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()


#define r register
#define ll long long

using namespace std;

const int N = 1e6 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e6 + 3;
const int p = 1e6 + 3;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n = 11;
LL a[20][20], ans[N], maxi, tmp;

ll ksm(r ll x,r int y)
{
    if(!y) return 1;
    r ll ret=ksm(x,y>>1);
    if(y&1) return ret*ret%p*x%p;
    return ret*ret%p;
}

int Ask(int x) { printf("? %d
", x); fflush(stdout); int y; scanf("%d", &y); return y;}
// 偷的
int main() {
    for(int i = 1; i <= n; i++) {
        LL val = Ask(i - 1);
        for(int j = 1; j <= n; j++) a[i][j] = ksm(i - 1, j - 1);
        a[i][n + 1] = val;
    }
    for(r int i=1;i<=n;i++)
    {
        if(!a[i][i])//主元不能为0
        {
            maxi=0;
            for(r int j=i+1;j<=n&&!maxi;j++)
                if(a[j][i]) maxi=j;
            if(!maxi) continue;//如果一整列都为0，不需要消元
            for(r int j=i;j<=n+1;j++)
                tmp=a[maxi][j],a[maxi][j]=a[i][j],a[i][j]=tmp;
        }
        for(r int j=i+1;j<=n;j++)
        {
            tmp=a[j][i];
            if(!tmp) continue;//已经为0，不需要消元
            for(r int k=i;k<=n+1;k++)
                a[j][k]=((a[j][k]*a[i][i]-a[i][k]*tmp)%p+p)%p;
        }
    }
    for(r int i=n;i;i--)
    {
        for(r int j=i+1;j<=n;j++)
            a[i][n+1]=((a[i][n+1]-ans[j]*a[i][j])%p+p)%p;
        ans[i]=a[i][n+1]*ksm(a[i][i],p-2)%p;
    }
    for(int i = 0; i < p; i++) {
        LL ret = 0;
        for(int j = 0; j < 11; j++)
            ret = (ret + ksm(i, j) * ans[j + 1] % mod) % mod;
        if(ret==0) {
            printf("! %d
", i);
            fflush(stdout);
            return 0;
        }
    }
    puts("! -1");
    fflush(stdout);
    return 0;
}
#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()


#define r register
#define ll long long

using namespace std;

const int N = 1e6 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e6 + 3;
const int p = 1e6 + 3;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n = 11;
LL a[20][20], ans[N], maxi, tmp;

ll ksm(r ll x,r int y)
{
    if(!y) return 1;
    r ll ret=ksm(x,y>>1);
    if(y&1) return ret*ret%p*x%p;
    return ret*ret%p;
}

int Ask(int x) { printf("? %d
", x); fflush(stdout); int y; scanf("%d", &y); return y;}
// 偷的
int main() {
    for(int i = 1; i <= n; i++) {
        LL val = Ask(i - 1);
        for(int j = 1; j <= n; j++) a[i][j] = ksm(i - 1, j - 1);
        a[i][n + 1] = val;
    }
    for(r int i=1;i<=n;i++)
    {
        if(!a[i][i])//主元不能为0
        {
            maxi=0;
            for(r int j=i+1;j<=n&&!maxi;j++)
                if(a[j][i]) maxi=j;
            if(!maxi) continue;//如果一整列都为0，不需要消元
            for(r int j=i;j<=n+1;j++)
                tmp=a[maxi][j],a[maxi][j]=a[i][j],a[i][j]=tmp;
        }
        for(r int j=i+1;j<=n;j++)
        {
            tmp=a[j][i];
            if(!tmp) continue;//已经为0，不需要消元
            for(r int k=i;k<=n+1;k++)
                a[j][k]=((a[j][k]*a[i][i]-a[i][k]*tmp)%p+p)%p;
        }
    }
    for(r int i=n;i;i--)
    {
        for(r int j=i+1;j<=n;j++)
            a[i][n+1]=((a[i][n+1]-ans[j]*a[i][j])%p+p)%p;
        ans[i]=a[i][n+1]*ksm(a[i][i],p-2)%p;
    }
    for(int i = 0; i < p; i++) {
        LL ret = 0;
        for(int j = 0; j < 11; j++)
            ret = (ret + ksm(i, j) * ans[j + 1] % mod) % mod;
        if(ret==0) {
            printf("! %d
", i);
            fflush(stdout);
            return 0;
        }
    }
    puts("! -1");
    fflush(stdout);
    return 0;
}