POJ1288 Sly Number(高斯消元 dfs枚举)

由于解集只为{0， 1， 2}故消元后需dfs枚举求解

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 60, INF = 0x3F3F3F3F;

int a[N][N], mod;
int x[N];
int n, row;
int ans[N];
bool ok;

int gauss(int a[][N], int n){
	int i, j;
	for(i = 0, j = 0; i < n && j < n; i++, j++){
		int r = i;
		for(int k = i; k < n; k++){
			if(a[k][j]){
				r = k;
				break;
			}
		}
		if(a[r][j] == 0){
			i--;
			continue;
		}
		if(r != i){
			for(int k = 0; k <= n; k++){
				swap(a[i][k], a[r][k]);
			}
		}
		for(int k = i + 1; k < n; k++){
			if(a[k][j]){
				int x1 = a[i][j], x2 = a[k][j];
				for(int l = j; l <= n; l++){
					a[k][l] = (a[k][l] * x1 - x2 * a[i][l]) % mod;
				}
			}
		}
	}
	return i;
}

void dfs(int r){
	if(r == -1){
		ok = 1;
		return;
	}
	if(ok){
		return;
	}
	int x = 0;
	while(x < n && a[r][x] == 0){
		x++;
	}
	if(x == n){
		if(a[r][n]){
			return;
		}
		for(ans[x] = 0; ans[x] <= 2; ans[x]++){
			dfs(r - 1);
		}
		return;
	}
    int tp = 0;
	for(int j = x + 1; j < n; j++){
		tp += a[r][j] * ans[j];
		tp %= mod;
	}
	for(ans[x] = 0; ans[x] <= 2; ans[x]++){
		if((ans[x] * a[r][x] + tp - a[r][n]) % mod == 0){
			dfs(r - 1);
		}
	}
}

int main(){
    int t;
    cin>>t;
    while(t--){
    	cin >> mod >> n;
    	for(int i = 0; i < n; i++){
    		cin >> x[i];
    	}
    	for(int i = 0; i < n; i++){
    		a[i][n] = (i == 0);
    		for(int j = 0, k = i; j <= i; j++ , k--){
    			a[i][j] = x[k];
    		}
    		for(int j = i + 1, k = n -1; j < n; j++, k--){
    			a[i][j] = x[k];
    		}
    	}
    	row = gauss(a, n);
    	ok = 0;
    	dfs(n - 1);
    	if(ok){
    		printf("A solution can be found\n");
    	}else{
    		printf("No solution\n");
    	}

    }

    return 0;
}