USACO 6.4 The Primes

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

 /*
 LANG: C++
 PROG: prime3
 */
 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #define siji(i,x,y) for(int i=(x); i<=(y) ; ++i)
 #define gongzi(j,x,y) for(int j=(x); j>=(y) ; --j)
 #define ivorysi
 using namespace std;
 int n,fi;
 int p[][],sum,prime[],tot,all;
 bool noprime[];
 typedef struct {
     int l,r[];
 }RECORD200;
 /*typedef struct {
     int l,r[50];
 }RECORD50;*/
 typedef struct {
     int l,r[];
 }RECORD20;
 //.0-9 _ 1.3.7.9  - non-zero
 RECORD200 pat1[];//X..._ 4*10*10
 RECORD200 pat2[];//_.X._ 4*4*10
 RECORD20 pat3[][];//X___Y 4*4
 RECORD20 pat4[][][];//-X.YZ 9
 RECORD20 pat7[][][];//X.Y.Z 10
 RECORD20 pat8[][][];//.XYZ_ 4
 string sol[];
 int marks[][];
 bool check(int x) {
     if(x%== || x%==) return ;
     return ;
 }
 void makeprime() {
     siji(i,,) {
         if(!noprime[i]) {
             prime[++tot]=i;
             if(i>) {
                 int temp=;
                 for(int k=;k<=;k*=) {
                     temp+=i/k%;
                 }
                 if(temp==n) {
                     ++sum;
                     for(int k=,m=;k<= && m>=;k*=,--m) {
                         p[sum][m]=i/k%;
                     }
                     int t1,t2,t3,t4,t5;
                     t1=p[sum][],t2=p[sum][],t3=p[sum][],t4=p[sum][],t5=p[sum][];
                     pat1[t1].r[++pat1[t1].l]=sum;
                     if(check(t1)) pat2[t3].r[++pat2[t3].l]=sum;
                     if(check(t2)&&check(t3)&&check(t4)) {
                         pat3[t1][t5].r[++pat3[t1][t5].l]=sum;
                     }
                     pat4[t2][t4][t5].r[++pat4[t2][t4][t5].l]=sum;
                     pat7[t1][t3][t5].r[++pat7[t1][t3][t5].l]=sum;
                     pat8[t2][t3][t4].r[++pat8[t2][t3][t4].l]=sum;
                 }
             }
         }
         for(int j=;j<=tot && i</prime[j];++j) {
             noprime[prime[j]*i]=;
             if(i%prime[j]==) break;
         }
     }
 }
 void addsolution() {
     ++all;
     siji(i,,) {
         siji(j,,) {
             sol[all].append(,marks[i][j]+'');
         }
     }
 }
 int num(int x) {
     int res=;
     siji(i,,) res=res*+p[x][i];
     return res;
 }
 void solve() {
     scanf("%d%d",&n,&fi);
     makeprime();
     int us,tj,tk,tm,tw,ty,tx,tq,temp,temp1,temp2;
     for(int i=;i<=pat1[fi].l;++i) {
         memset(marks,,sizeof(marks));
         us=pat1[fi].r[i];
         siji(iv,,) marks[iv][iv]=p[us][iv];
         tj=pat2[marks[][]].l;
         for(int j=;j<=tj;++j) {
             us=pat2[marks[][]].r[j];
             siji(iv,,) marks[-iv+][iv]=p[us][iv];
             tk=pat3[marks[][]][marks[][]].l;
             for(int k=;k<=tk;++k) {
                 us=pat3[marks[][]][marks[][]].r[k];
                 siji(iv,,) marks[][iv]=p[us][iv];
                 tm=pat4[marks[][]][marks[][]][marks[][]].l;
                 for(int m=;m<=tm;++m) {
                     us=pat4[marks[][]][marks[][]][marks[][]].r[m];
                     marks[][]=p[us][];
                     marks[][]=p[us][];
                     tw=pat4[marks[][]][marks[][]][marks[][]].l;
                     for(int w=;w<=tw;++w) {
                         us=pat4[marks[][]][marks[][]][marks[][]].r[w];
                         marks[][]=p[us][];
                         marks[][]=p[us][];
                         marks[][]=;
                         temp=;
                         siji(iv,,) temp+=marks[][iv];
                         marks[][]=n-temp;
                         if(marks[][]< || marks[][]>) continue;
                         temp=;
                         siji(iv,,) {
                             temp=temp*+marks[][iv];
                         }
                         if(noprime[temp]) continue;
                         ty=pat7[marks[][]][marks[][]][marks[][]].l;
                         for(int y=;y<=ty;++y) {
                             us=pat7[marks[][]][marks[][]][marks[][]].r[y];
                             marks[][]=p[us][];
                             marks[][]=p[us][];
                             tx=pat8[marks[][]][marks[][]][marks[][]].l;
                             for(int x=;x<=tx;++x) {
                                 us=pat8[marks[][]][marks[][]][marks[][]].r[x];
                                 marks[][]=p[us][];
                                 marks[][]=p[us][];
                                 tq=pat8[marks[][]][marks[][]][marks[][]].l;
                                 for(int q=;q<=tq;++q) {
                                     us=pat8[marks[][]][marks[][]][marks[][]].r[q];
                                     marks[][]=p[us][];
                                     marks[][]=p[us][];
                                     marks[][]=marks[][]=;
                                     temp1=temp2=;
                                     siji(iv,,) temp1+=marks[iv][],temp2+=marks[iv][];
                                     marks[][]=n-temp1;marks[][]=n-temp2;
                                     if(marks[][]< || marks[][]< ||marks[][]> || marks[][]>)
                                         continue;
                                     temp1=temp2=;
                                     siji(iv,,)
                                         temp1=temp1*+marks[iv][],temp2=temp2*+marks[iv][];
                                     if(noprime[temp1]||noprime[temp2]) continue;
                                     temp1=temp2=;
                                     siji(iv,,)
                                         temp1=temp1*+marks[][iv],temp2+=marks[][iv];
                                     if(noprime[temp1] || temp2!=n) continue;
                                     addsolution();
                                 }
                             }
                         }
                     }
                 }
             }
         }
     }
     if(all==) puts("NONE");
     sort(sol+,sol+all+);
     siji(i,,all) {
         for(int j=;j<;++j) {
             cout<<sol[i].substr(j*,)<<endl;
         }
         if(i!=all) puts("");
     }
 }
 int main(int argc, char const *argv[])
 {
 #ifdef ivorysi
     freopen("prime3.in","r",stdin);
     freopen("prime3.out","w",stdout);
 #else
     freopen("f1.in","r",stdin);
     freopen("f1.out","w",stdout);
 #endif
     solve();
     return ;
 }