USACO 6.4 The Primes

Posted 2020-09-20 sigongzi

The Primes
IOI‘94

In the square below, each row, each column and the two diagonals can be read as a five digit prime number. The rows are read from left to right. The columns are read from top to bottom. Both diagonals are read from left to right.

+---+---+---+---+---+
| 1 | 1 | 3 | 5 | 1 |
+---+---+---+---+---+
| 3 | 3 | 2 | 0 | 3 |
+---+---+---+---+---+
| 3 | 0 | 3 | 2 | 3 |
+---+---+---+---+---+
| 1 | 4 | 0 | 3 | 3 |
+---+---+---+---+---+
| 3 | 3 | 3 | 1 | 1 |
+---+---+---+---+---+ 

• The prime numbers‘ digits must sum to the same number.
• The digit in the top left-hand corner of the square is pre-determined (1 in the example).
• A prime number may be used more than once in the same square.
• If there are several solutions, all must be presented (sorted in numerical order as if the 25 digits were all one long number).
• A five digit prime number cannot begin with a zero (e.g., 00003 is NOT a five digit prime number).

PROGRAM NAME: prime3

INPUT FORMAT

A single line with two space-separated integers: the sum of the digits and the digit in the upper left hand corner of the square.

SAMPLE INPUT (file prime3.in)

11 1

OUTPUT FORMAT

Five lines of five characters each for each solution found, where each line in turn consists of a five digit prime number. Print a blank line between solutions. If there are no prime squares for the input data, output a single line containing "NONE".

SAMPLE OUTPUT (file prime3.out)

The above example has 3 solutions.

11351
14033
30323
53201
13313

11351
33203
30323
14033
33311

13313
13043
32303
50231
13331

————————————————————————题解
其实这又是一道搜索顺序至关重要的搜索题
计算机比人脑强大的地方就是能做大量重复的复杂运算
我们发现主对角线其实影响最大，然后是次对角线，然后逐渐再将限制较多的行或列填上，例如最后一行或者最后一列的数必须是1,3,7,9
预处理出所有5位、各数位的和为n的素数。
按照这样的搜索顺序
1　　4　　6　　5　　2
8　　1　　7　　2　　8
10　　4　　1　　5　　10
9　　2　　7　　1　　9
2　　3　　3　　3　　1
这样我们可以就可以少打几个循环了……
在计算素数的时候同时求一些一些数位固定的数
【其中XYZ为已知数，.为未知数，_ 为1，3,7,9】
例如枚举1，X ... _
枚举2，_ . X . _
枚举3，X_ _ _ Y
枚举4,5 .X . Y Z
6可以直接计算
枚举7 X . Y . Z
枚举8，9 . X Y Z _
10可以直接计算
然后时限为2s的题就可以0.011过了

  1 /*
  2 LANG: C++
  3 PROG: prime3
  4 */
  5 #include <iostream>
  6 #include <cstdio>
  7 #include <algorithm>
  8 #include <cstring>
  9 #define siji(i,x,y) for(int i=(x); i<=(y) ; ++i)
 10 #define gongzi(j,x,y) for(int j=(x); j>=(y) ; --j)
 11 #define ivorysi
 12 using namespace std;
 13 int n,fi;
 14 int p[1005][6],sum,prime[50005],tot,all;
 15 bool noprime[100005];
 16 typedef struct {
 17     int l,r[200];
 18 }RECORD200;
 19 /*typedef struct {
 20     int l,r[50];
 21 }RECORD50;*/
 22 typedef struct {
 23     int l,r[20];
 24 }RECORD20;
 25 //.0-9 _ 1.3.7.9  - non-zero
 26 RECORD200 pat1[10];//X..._ 4*10*10
 27 RECORD200 pat2[10];//_.X._ 4*4*10
 28 RECORD20 pat3[10][10];//X___Y 4*4
 29 RECORD20 pat4[10][10][10];//-X.YZ 9
 30 RECORD20 pat7[10][10][10];//X.Y.Z 10
 31 RECORD20 pat8[10][10][10];//.XYZ_ 4
 32 string sol[1005];
 33 int marks[6][6];
 34 bool check(int x) {
 35     if(x%2==0 || x%5==0) return 0;
 36     return 1;
 37 }
 38 void makeprime() {
 39     siji(i,2,99999) {
 40         if(!noprime[i]) {
 41             prime[++tot]=i;
 42             if(i>10000) {
 43                 int temp=0;
 44                 for(int k=1;k<=10000;k*=10) {
 45                     temp+=i/k%10;
 46                 }
 47                 if(temp==n) {
 48                     ++sum;
 49                     for(int k=1,m=5;k<=10000 && m>=1;k*=10,--m) {
 50                         p[sum][m]=i/k%10;
 51                     }
 52                     int t1,t2,t3,t4,t5;
 53                     t1=p[sum][1],t2=p[sum][2],t3=p[sum][3],t4=p[sum][4],t5=p[sum][5];
 54                     pat1[t1].r[++pat1[t1].l]=sum;    
 55                     if(check(t1)) pat2[t3].r[++pat2[t3].l]=sum;
 56                     if(check(t2)&&check(t3)&&check(t4)) {
 57                         pat3[t1][t5].r[++pat3[t1][t5].l]=sum;
 58                     }
 59                     pat4[t2][t4][t5].r[++pat4[t2][t4][t5].l]=sum;
 60                     pat7[t1][t3][t5].r[++pat7[t1][t3][t5].l]=sum;
 61                     pat8[t2][t3][t4].r[++pat8[t2][t3][t4].l]=sum;
 62                 }
 63             }
 64         }
 65         for(int j=1;j<=tot && i<100000/prime[j];++j) {
 66             noprime[prime[j]*i]=1;
 67             if(i%prime[j]==0) break; 
 68         } 
 69     }
 70 }
 71 void addsolution() {
 72     ++all;
 73     siji(i,1,5) {
 74         siji(j,1,5) {
 75             sol[all].append(1,marks[i][j]+‘0‘);
 76         }
 77     }
 78 }
 79 int num(int x) {
 80     int res=0;
 81     siji(i,1,5) res=res*10+p[x][i];
 82     return res;
 83 }
 84 void solve() {
 85     scanf("%d%d",&n,&fi);
 86     makeprime();
 87     int us,tj,tk,tm,tw,ty,tx,tq,temp,temp1,temp2;
 88     for(int i=1;i<=pat1[fi].l;++i) {
 89         memset(marks,0,sizeof(marks));
 90         us=pat1[fi].r[i];
 91         siji(iv,1,5) marks[iv][iv]=p[us][iv];
 92         tj=pat2[marks[3][3]].l;
 93         for(int j=1;j<=tj;++j) {
 94             us=pat2[marks[3][3]].r[j];
 95             siji(iv,1,5) marks[5-iv+1][iv]=p[us][iv];
 96             tk=pat3[marks[5][1]][marks[5][5]].l;
 97             for(int k=1;k<=tk;++k) {
 98                 us=pat3[marks[5][1]][marks[5][5]].r[k];
 99                 siji(iv,2,4) marks[5][iv]=p[us][iv];
100                 tm=pat4[marks[2][2]][marks[4][2]][marks[5][2]].l;
101                 for(int m=1;m<=tm;++m) {
102                     us=pat4[marks[2][2]][marks[4][2]][marks[5][2]].r[m];
103                     marks[1][2]=p[us][1];
104                     marks[3][2]=p[us][3];
105                     tw=pat4[marks[2][4]][marks[4][4]][marks[5][4]].l;
106                     for(int w=1;w<=tw;++w) {
107                         us=pat4[marks[2][4]][marks[4][4]][marks[5][4]].r[w];
108                         marks[1][4]=p[us][1];
109                         marks[3][4]=p[us][3];
110                         marks[1][3]=0;
111                         temp=0;
112                         siji(iv,1,5) temp+=marks[1][iv];
113                         marks[1][3]=n-temp;
114                         if(marks[1][3]<1 || marks[1][3]>9) continue;
115                         temp=0;
116                         siji(iv,1,5) {
117                             temp=temp*10+marks[1][iv];
118                         }
119                         if(noprime[temp]) continue;
120                         ty=pat7[marks[1][3]][marks[3][3]][marks[5][3]].l;
121                         for(int y=1;y<=ty;++y) {
122                             us=pat7[marks[1][3]][marks[3][3]][marks[5][3]].r[y];
123                             marks[2][3]=p[us][2];
124                             marks[4][3]=p[us][4];
125                             tx=pat8[marks[2][2]][marks[2][3]][marks[2][4]].l;
126                             for(int x=1;x<=tx;++x) {
127                                 us=pat8[marks[2][2]][marks[2][3]][marks[2][4]].r[x];
128                                 marks[2][1]=p[us][1];
129                                 marks[2][5]=p[us][5];
130                                 tq=pat8[marks[4][2]][marks[4][3]][marks[4][4]].l;
131                                 for(int q=1;q<=tq;++q) {
132                                     us=pat8[marks[4][2]][marks[4][3]][marks[4][4]].r[q];
133                                     marks[4][1]=p[us][1];
134                                     marks[4][5]=p[us][5];
135                                     marks[3][1]=marks[3][5]=0;
136                                     temp1=temp2=0;
137                                     siji(iv,1,5) temp1+=marks[iv][1],temp2+=marks[iv][5];
138                                     marks[3][1]=n-temp1;marks[3][5]=n-temp2;
139                                     if(marks[3][1]<1 || marks[3][5]<1 ||marks[3][1]>9 || marks[3][5]>9) 
140                                         continue;
141                                     temp1=temp2=0;
142                                     siji(iv,1,5) 
143                                         temp1=temp1*10+marks[iv][1],temp2=temp2*10+marks[iv][5];
144                                     if(noprime[temp1]||noprime[temp2]) continue;
145                                     temp1=temp2=0;
146                                     siji(iv,1,5) 
147                                         temp1=temp1*10+marks[3][iv],temp2+=marks[3][iv];
148                                     if(noprime[temp1] || temp2!=n) continue;
149                                     addsolution();
150                                 }
151                             }
152                         }
153                     }
154                 }
155             }
156         }
157     }
158     if(all==0) puts("NONE");
159     sort(sol+1,sol+all+1);
160     siji(i,1,all) {
161         for(int j=0;j<5;++j) {
162             cout<<sol[i].substr(j*5,5)<<endl;
163         }
164         if(i!=all) puts("");
165     }
166 }
167 int main(int argc, char const *argv[])
168 {
169 #ifdef ivorysi
170     freopen("prime3.in","r",stdin);
171     freopen("prime3.out","w",stdout);
172 #else
173     freopen("f1.in","r",stdin);
174     freopen("f1.out","w",stdout);
175 #endif
176     solve();
177     return 0;
178 }