题目大意
给一个5*5的矩阵,在每个格子里面填入0~9这10个数字.要求行从左往右读,列从上往下读,两条对角线从上往下读是一个5位素数且不含前导0.
现在给出了开头的数字和行,列,对角线之和,求符合规定的所有数字矩阵.
题解
这是一道搜索题.
但盲目的搜索是会TLE的.因此,我们应当采用合理的剪枝.
由题目可知,加入行列对角线我们知道了任意4位数,我们可以推出第5位数是什么.
因此,我们可以采用传说中的手动指定搜索顺序的方法来搜索和回溯.
具体搜索顺序如下图.
大概就是这样的感觉.
里面的C是可以直接算出来的.所以我们只需要搜索只有数字那些就行了.
还有我一开始是枚举对角线素数.
注意: 先枚举每个数的最后一位会让你减少至少一半的搜索量.因为5位素数的最后一位数肯定不是偶数.
详情请看代码.代码有点恶心.我用了宏定义,因为这样看起来会更加恶心.
代码
/*
TASK:prime3
LANG:C++
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#define REP(i,st,ed) for ((i) = (st); (i) < (ed); ++(i))
using namespace std;
struct Ans
{
int matrix[5][5];
Ans operator = (Ans &rhs)
{
memcpy(matrix, rhs.matrix, sizeof(rhs.matrix));
return *this;
}
bool operator < (const Ans &rhs) const
{
for (int i = 0; i < 5; ++i)
for (int j = 0; j < 5; ++j)
if (matrix[i][j] < rhs.matrix[i][j]) return true;
else if (matrix[i][j] > rhs.matrix[i][j]) return false;
return false;
}
}ans[1000], tmp;
int anslen;
const int MAXP = 100000;
int p[10000], np, sumallr[5], sumallc[5], sumalld, sum;
bool v[MAXP * 10];
bool check(int x)
{
int s = 0;
while (x) s += x % 10, x /= 10;
return s == sum;
}
void filtrate()
{
memset(v, true, sizeof(v));
for (int i = 2; i < MAXP; ++i)
for (int j = 2; i * j < MAXP; ++j)
v[i * j] = false;
np = 0;
for (int i = 10000; i < MAXP; ++i)
if (v[i] && check(i)) p[np++] = i;
}
bool judge(int x, int y)
{
if (sumallr[x] + tmp.matrix[x][y] > sum) return false;
if (sumallc[y] + tmp.matrix[x][y] > sum) return false;
if (4 - x == y && sumalld + tmp.matrix[x][y] > sum) return false;
return true;
}
void adds(int x, int y, int d)
{
sumallr[x] += tmp.matrix[x][y] * d;
sumallc[y] += tmp.matrix[x][y] * d;
if (4 - x == y) sumalld += tmp.matrix[x][y] * d;
}
int trans(int d, int m)
{
if (m == 1) return tmp.matrix[d][0] * 10000
+ tmp.matrix[d][1] * 1000
+ tmp.matrix[d][2] * 100
+ tmp.matrix[d][3] * 10
+ tmp.matrix[d][4];
else if (m == 2) return tmp.matrix[0][d] * 10000
+ tmp.matrix[1][d] * 1000
+ tmp.matrix[2][d] * 100
+ tmp.matrix[3][d] * 10
+ tmp.matrix[4][d];
else if (m == 3) return tmp.matrix[4][0] * 10000
+ tmp.matrix[3][1] * 1000
+ tmp.matrix[2][2] * 100
+ tmp.matrix[1][3] * 10
+ tmp.matrix[0][4];
}
bool condit(int x)
{
return x >= 0 && x <= 9;
}
int main()
{
freopen("prime3.in", "r", stdin);
freopen("prime3.out", "w", stdout);
scanf("%d%d", &sum, &tmp.matrix[0][0]);
filtrate();
memset(sumallr, 0, sizeof(sumallr));
memset(sumallc, 0, sizeof(sumallc));
sumalld = 0;
anslen = 0;
for (int i = 0; i < np; ++i)
{
if (p[i] * 10 / MAXP != tmp.matrix[0][0]) continue;
int k = p[i];
for (int j = 4; j >= 0; --j)
{
tmp.matrix[j][j] = k % 10;
sumallr[j] = sumallc[j] = k % 10;
k /= 10;
}
sumalld = p[i] % 1000 / 100;
REP(tmp.matrix[0][4], 1, 10)
{
if (tmp.matrix[0][4] % 2 == 0 || tmp.matrix[0][4] == 5) continue;
if (!judge(0, 4)) break;
adds(0, 4, 1);
REP(tmp.matrix[1][3], 0, 10)
{
if (!judge(1, 3)) break;
adds(1, 3, 1);
REP(tmp.matrix[4][0], 1, 10)
{
if (tmp.matrix[4][0] % 2 == 0 || tmp.matrix[4][0] == 5) continue;
if (!judge(4, 0)) break;
tmp.matrix[3][1] = sum - sumalld - tmp.matrix[4][0];
if (!condit(tmp.matrix[3][1]) || !judge(3, 1)) continue;
if (!v[trans(0, 3)]) continue;
adds(3, 1, 1);
adds(4, 0, 1);
REP(tmp.matrix[1][4], 1, 10)
{
if (tmp.matrix[1][4] % 2 == 0 || tmp.matrix[1][4] == 5) continue;
if (!judge(1, 4)) break;
adds(1, 4, 1);
REP(tmp.matrix[2][4], 1, 10)
{
if (tmp.matrix[2][4] % 2 == 0 || tmp.matrix[2][4] == 5) continue;
if (!judge(2, 4)) break;
tmp.matrix[3][4] = sum - sumallc[4] - tmp.matrix[2][4];
if (!condit(tmp.matrix[3][4]) || !judge(3, 4)) continue;
if (!v[trans(4, 2)]) continue;
adds(3, 4, 1);
adds(2, 4, 1);
REP(tmp.matrix[4][3], 1, 10)
{
if (tmp.matrix[4][3] % 2 == 0 || tmp.matrix[4][3] == 5) continue;
if (!judge(4, 3)) break;
adds(4, 3, 1);
REP(tmp.matrix[0][3], 1, 10)
{
if (!judge(0, 3)) break;
tmp.matrix[2][3] = sum - sumallc[3] - tmp.matrix[0][3];
if (!condit(tmp.matrix[2][3]) || !judge(2, 3)) continue;
if (!v[trans(3, 2)]) continue;
adds(0, 3, 1);
adds(2, 3, 1);
REP(tmp.matrix[4][2], 1, 10)
{
if (tmp.matrix[4][2] % 2 == 0 || tmp.matrix[4][2] == 5) continue;
if (!judge(4, 2)) break;
adds(4, 2, 1);
REP(tmp.matrix[3][2], 0, 10)
{
if (!judge(3, 2)) break;
adds(3, 2, 1);
REP(tmp.matrix[0][2], 1, 10)
{
if (!judge(0, 2)) break;
tmp.matrix[1][2] = sum - sumallc[2] - tmp.matrix[0][2];
if (!condit(tmp.matrix[1][2]) || !judge(1, 2) || !v[trans(2, 2)]) continue;
tmp.matrix[0][1] = sum - sumallr[0] - tmp.matrix[0][2];
if (tmp.matrix[0][1] == 0) continue;
if (!condit(tmp.matrix[0][1]) || !judge(0, 1) || !v[trans(0, 1)]) continue;
tmp.matrix[4][1] = sum - sumallr[4];
if (!condit(tmp.matrix[4][1]) || !judge(4, 1) || !v[trans(4, 1)]) continue;
tmp.matrix[2][1] = sum - sumallc[1] - tmp.matrix[0][1] - tmp.matrix[4][1];
if (!condit(tmp.matrix[2][1]) || !judge(2, 1) || !v[trans(1, 2)]) continue;
tmp.matrix[1][0] = sum - sumallr[1] - tmp.matrix[1][2];
if (tmp.matrix[1][0] == 0) continue;
if (!condit(tmp.matrix[1][0]) || !judge(1, 0) || !v[trans(1, 1)]) continue;
tmp.matrix[2][0] = sum - sumallr[2] - tmp.matrix[2][1];
if (tmp.matrix[2][0] == 0) continue;
if (!condit(tmp.matrix[2][0]) || !judge(2, 0) || !v[trans(2, 1)]) continue;
if (sumallc[0] + tmp.matrix[1][0] + tmp.matrix[2][0] != sumallr[3]) continue;
tmp.matrix[3][0] = sum - sumallr[3];
if (tmp.matrix[3][0] == 0) continue;
if (!condit(tmp.matrix[3][0]) || !v[trans(0, 2)] || !v[trans(3, 1)]) continue;
ans[anslen++] = tmp;
}
adds(3, 2, -1);
}
adds(4, 2, -1);
}
adds(0, 3, -1);
adds(2, 3, -1);
}
adds(4, 3, -1);
}
adds(3, 4, -1);
adds(2, 4, -1);
}
adds(1, 4, -1);
}
adds(4, 0, -1);
adds(3, 1, -1);
}
adds(1, 3, -1);
}
adds(0, 4, -1);
}
}
sort(ans, ans + anslen);
for (int k = 0; k < anslen; ++k)
{
if (k != 0) printf("\n");
for (int i = 0; i < 5; ++i)
{
for (int j = 0; j < 5; ++j)
printf("%d", ans[k].matrix[i][j]);
printf("\n");
}
}
return 0;
}
时间： 2024-10-31 20:00:59