USACO4.3.2——质数方阵暴力解法

题面链接：这里；

读过题后，就不难发现这是一道填数题，先说填法，大多数人一开始肯定想到一行行填，一列列填，填完后整体判断——大数据直接把你T废：

于是我们就开始关注一些已经确定的点，确定的点越多，可供选择的素数就越少，运行速度就越快，请看下图：

图中的深红色（最左上角）表示题目中确定的点，我们先填第一行第一列（红色），这样有一个约束条件（一头确定），再填右上-左下的那条斜边（紫色），这样有两个约束条件（一头一尾已经确定），再填左上-右下的那条斜边（深蓝），再填浅蓝和绿色，它们都有两个约束条件。最后算出橙色的值，用总值减去其他四个数的值，如果是负数或大于10，就退回来重新做。

整体做完后，按照题目所给的条件进行判断，不符合就退回来重新做。

输出答案前需要先排序，将答案二维数组存到一个字符串中，这样就可以方便比较。

这里需要重点提的一点便是判质，下面是我刚开始写的一段代码：

bool Judge_Num(int Num) {for (register int i = 2; i <= sqrt(Num); i++)if (Num % i == 0)return false;// cout<<Num<<endl;return true;
}

注意，它会爆——每次循环计算一次sqrt是很耗时间的，所以，我做了以下改进：

bool Judge_Num(int Num) {int T=sqrt(Num);//这样只要计算一次sqrtfor (register int i = 2; i <= T; i++)if (Num % i == 0)return false;// cout<<Num<<endl;return true;
}

可是还是会超时（Num是一个五位数，每次循环少说也要100次），能不能将判质时间缩到O(1)呢？
可以的：
我们在填数之前先用欧拉筛将10000到99999中的素数筛出来，用bool数组Visited储存，Visited[N]=1表示N是质数，Visited[N]=0表示N是不是质数，这样，每次判质时，我们只需要：

bool Judge_Num(int Num){return true-Visited[Num];
}

万事大吉

代码（号称USACO史上最长）：

#include "bits/stdc++.h"
using namespace std;struct Unsigned {int Size, Num[1001];int E[1001][11];
} Prime[11];//储存每一个素数
int Sum, Start, Used[9], Result[9][9], Size, prime[100100];
bool Can, Visited[100101];
const int N = 5, M = 9;struct Out {string Name;int Num_Result[9][9];friend bool operator<(Out Num1, Out Num2) { return Num1.Name < Num2.Name; }
} HaHaHa[999];
//答案的排序void Made_Prime(void) {for (int i = 2; i <= 100100; i++) {if (!Visited[i])prime[++prime[0]] = i;for (int j = 1; j <= prime[0] && i * prime[j] <= 100100; j++) {Visited[i * prime[j]] = 1;if (i % prime[j] == 0)break;}}return;
}//欧拉筛bool Judge_Num(int Num) { return true - Visited[Num];
}//判质void DFS_Made(int K, int Last, int Already) {if (Last > 9 * (N - K))return;if (K == N) {if (Last == 0 && Judge_Num(Already) == true) {int JJ = Already, KK = N;++Prime[Already / 10000].Size;int UU = Prime[Already / 10000].Size;Prime[Already / 10000].Num[UU] = Already;while (JJ != 0) {Prime[Already / 10000].E[UU][KK] = JJ % 10;KK--, JJ = JJ / 10;}}return;}for (register int i = 0; i <= 9; i++) DFS_Made(K + 1, Last - i, Already * 10 + i);return;
}//将合法五位素数整出来bool Judge(void) {int Sum_Judge = 0, Mul_Judge = 0;for (register int i = 1; i <= N; i++) {Sum_Judge = 0;for (register int j = 1; j <= N; j++) Sum_Judge = Sum_Judge + Result[j][i];if (Sum_Judge != Sum)return false;}for (register int i = 1; i <= N; i++) {Mul_Judge = 0;for (register int j = 1; j <= N; j++) Mul_Judge = Mul_Judge * 10 + Result[j][i];if (Judge_Num(Mul_Judge) == false)return false;}for (register int i = 1; i <= N; i++) {Sum_Judge = 0;for (register int j = 1; j <= N; j++) Sum_Judge = Sum_Judge + Result[i][j];if (Sum_Judge != Sum)return false;}for (register int i = 1; i <= N; i++) {Mul_Judge = 0;for (register int j = 1; j <= N; j++) Mul_Judge = Mul_Judge * 10 + Result[i][j];if (Judge_Num(Mul_Judge) == false)return false;}Sum_Judge = 0, Mul_Judge = 0;for (register int i = 1; i <= N; i++) {Mul_Judge = Mul_Judge * 10 + Result[i][i];Sum_Judge = Sum_Judge + Result[i][i];}if (Sum_Judge != Sum || Judge_Num(Mul_Judge) == false)return false;Sum_Judge = 0, Mul_Judge = 0;for (register int i = 1; i <= N; i++) {Mul_Judge = Mul_Judge * 10 + Result[N - i + 1][i];Sum_Judge = Sum_Judge + Result[N - i + 1][i];}if (Sum_Judge != Sum || Judge_Num(Mul_Judge) == false)return false;return true;
}bool Judge_Zero(int *QQ) {for (register int i = 1; i <= N; i++)if (QQ[i] == 0)return false;return true;
}//对矩阵的合法性进行判断int main(void) {cin >> Sum >> Start;if (Sum < Start + 4 || Sum % 3 == 0) {cout << "NONE" << endl;return 0;}//如果总和太小或每个五位数是三的倍数，直接无解Made_Prime();for (register int i = 1; i <= 9; i++) DFS_Made(1, Sum - i, i);
//将素数整出来
//以下代码比较长，其实就是将数填进格子中for (register int ii = 1; ii <= Prime[Start].Size; ii++) {if (Judge_Zero(Prime[Start].E[ii]) == true) {for (register int i = 1; i <= N; i++) Result[1][i] = Prime[Start].E[ii][i];for (register int iii = 1; iii <= Prime[Start].Size; iii++) {if (Judge_Zero(Prime[Start].E[iii]) == true) {for (register int i = 1; i <= N; i++) Result[i][1] = Prime[Start].E[iii][i];int T = Result[N][1];for (register int i = 1; i <= Prime[T].Size; i++) {if (Prime[T].E[i][N] == Result[1][N]) {for (register int j = 1; j <= N; j++) Result[N - j + 1][j] = Prime[T].E[i][j];int Y = Result[3][1];for (register int j = 1; j <= Prime[Y].Size; j++) {if (Prime[Y].E[j][3] == Result[3][3]) {for (register int k = 1; k <= N; k++) Result[3][k] = Prime[Y].E[j][k];int D = Result[1][1];for (register int k = 1; k <= Prime[D].Size; k++) {if (Prime[D].E[k][3] == Result[3][3]) {for (register int l = 1; l <= N; l++)Result[l][l] = Prime[D].E[k][l];Result[N][2] = Sum - Result[1][2] - Result[2][2] - Result[3][2] -Result[4][2];if (Result[N][2] < 0 || Result[N][2] > 9)continue;int JJJJJJJ = 0, KKKKKKK = 1;while (KKKKKKK <= N) {JJJJJJJ = JJJJJJJ * 10 + Result[KKKKKKK][2];KKKKKKK++;}if (Judge_Num(JJJJJJJ) == false)continue;//填到一半的一个小剪枝int G = Result[1][3];for (register int l = 1; l <= Prime[G].Size; l++) {if (Prime[G].E[l][3] == Result[3][3]) {for (register int h = 1; h <= N; h++)Result[h][3] = Prime[G].E[l][h];Result[2][N] = Sum - Result[2][1] - Result[2][2] -Result[2][3] - Result[2][4];Result[4][N] = Sum - Result[4][1] - Result[4][2] -Result[4][3] - Result[4][4];Result[N][2] = Sum - Result[1][2] - Result[2][2] -Result[3][2] - Result[4][2];Result[N][4] = Sum - Result[1][4] - Result[2][4] -Result[3][4] - Result[4][4];if (Result[2][N] < 0 || Result[4][N] < 0 ||Result[N][2] < 0 || Result[N][4] < 0)continue;if (Result[2][N] > 9 || Result[4][N] > 9 ||Result[N][2] > 9 || Result[N][4] > 9)continue;
//合法性判断if (Judge() == true) {Can = true;Size++;for (register int iiii = 1; iiii <= N; iiii++) {for (register int jj = 1; jj <= N; jj++) {HaHaHa[Size].Num_Result[iiii][jj] =Result[iiii][jj];HaHaHa[Size].Name += ('0' + Result[iiii][jj]);}  // cout<<endl;}      // cout<<endl;}}}}}}}}}}}}}if (Can == false) {cout << "NONE" << endl;} else {sort(HaHaHa + 1, HaHaHa + 1 + Size);for (register int k = 1; k <= Size; k++) {for (register int i = 1; i <= N; i++) {for (register int j = 1; j <= N; j++) {cout << HaHaHa[k].Num_Result[i][j];}//排序并输出cout << endl;}cout << endl;}}return 0;
}

The End