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 28 1 7 2 810 4 1 5 109 2 7 1 92 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 Z6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 }
时间： 2024-12-11 13:53:43