直接用最笨的方法。。。
ANALYSIS里有讲，其实只算一半长度就行
代码相当的难看。。。
1 /*
2 ID:linyvxi1
3 PROB:pprime
4 LANG:C++
5 */
6 #include <stdio.h>
7 #include <math.h>
8 void do_1();
9 void do_2();
10 void do_3();
11 void do_4();
12 void do_5();
13 void do_6();
14 void do_7();
15 void do_8();
16 inline bool prime(int n)
17 {
18 int i;
19 for(i=2;i<=(int)sqrt(n)+1;i++){
20 if(n%i==0)
21 return false;
22 }
23 return true;
24 }
25 int left,right,min_len,max_len;
26 void get_len()
27 {
28 min_len=0;
29 max_len=0;
30 int temp1=left;
31 int temp2=right;
32 while(left){
33 min_len++;
34 left/=10;
35 }
36 while(right){
37 max_len++;
38 right/=10;
39 }
40 left=temp1;
41 right=temp2;
42 }
43 void do_1(){
44 int cur_len=1;
45 int num1;
46 int palindromes;
47 for(num1=1;num1<=9;num1++){
48 palindromes=num1;
49 if(palindromes<left||!(prime(palindromes)))
50 continue;
51 if(palindromes>right)
52 return;
53 printf("%d\n",palindromes);
54 }
55 if(cur_len<max_len)
56 do_2();
57 }
58 void do_2()
59 {
60 int cur_len=2;
61 int num1;
62 int palindromes;
63 for(num1=1;num1<=9;num1+=2){
64 palindromes=11*num1;
65 if(palindromes<left||!(prime(palindromes)))
66 continue;
67 if(palindromes>right)
68 return;
69 printf("%d\n",palindromes);
70 }
71 if(cur_len<max_len)
72 do_3();
73 }
74 void do_3()
75 {
76 int cur_len=3;
77 int num1,num2;
78 int palindromes;
79 for(num1=1;num1<=9;num1+=2){
80 for(num2=0;num2<=9;num2++){
81 palindromes=101*num1+10*num2;
82 if(palindromes<left||!(prime(palindromes)))
83 continue;
84 if(palindromes>right)
85 return;
86 printf("%d\n",palindromes);
87 }
88 }
89 if(cur_len<max_len)
90 do_4();
91 }
92 void do_4()
93 {
94 int cur_len=4;
95 int num1,num2;
96 int palindromes;
97 for(num1=1;num1<=9;num1+=2){
98 for(num2=0;num2<=9;num2++){
99 palindromes=num1*1001*110*num2;
100 if(palindromes<left||!(prime(palindromes)))
101 continue;
102 if(palindromes>right)
103 return;
104 printf("%d\n",palindromes);
105 }
106 }
107 if(cur_len<max_len)
108 do_5();
109 }
110 void do_5()
111 {
112 int cur_len=5;
113 int num1,num2,num3;
114 int palindromes;
115 for(num1=1;num1<=9;num1+=2){
116 for(num2=0;num2<=9;num2++){
117 for(num3=0;num3<=9;num3++){
118 palindromes=10001*num1+1010*num2+num3*100;
119 if(palindromes<left||!prime(palindromes))
120 continue;
121 if(palindromes>right)
122 return;
123 printf("%d\n",palindromes);
124 }
125 }
126 }
127 if(cur_len<max_len)
128 do_6();
129 }
130 void do_6(){
131 int cur_len=6;
132 int num1,num2,num3;
133 int palindromes;
134 for(num1=1;num1<=9;num1++){
135 for(num2=0;num2<=9;num2++){
136 for(num3=0;num3<=9;num3++){
137 palindromes=100001*num1+10010*num2+1100*num3;
138 if(palindromes<left||!prime(palindromes))
139 continue;
140 if(palindromes>right)
141 return;
142 printf("%d\n",palindromes);
143 }
144 }
145 }
146 if(cur_len<max_len)
147 do_7();
148 }
149 void do_7()
150 {
151 int cur_len=7;
152 int num1,num2,num3,num4;
153 int palindromes;
154 for(num1=1;num1<=9;num1+=2){
155 for(num2=0;num2<=9;num2++){
156 for(num3=0;num3<=9;num3++){
157 for(num4=0;num4<=9;num4++){
158 palindromes=1000001*num1+100010*num2+10100*num3+1000*num4;
159 if(palindromes<left||!prime(palindromes))
160 continue;
161 if(palindromes>right)
162 return;
163 printf("%d\n",palindromes);
164 }
165 }
166 }
167 }
168 if(cur_len<max_len)
169 do_8();
170 }
171
172 void do_8()
173 {
174 int cur_len=8;
175 int num1,num2,num3,num4;
176 int palindromes;
177 for(num1=1;num1<=9;num1+=2){
178 for(num2=0;num2<=9;num2++){
179 for(num3=0;num3<=9;num3++){
180 for(num4=0;num4<=9;num4++){
181 palindromes=10000001*num1+1000010*num2+100100*num3+11000*num4;
182 if(palindromes<left||!prime(palindromes))
183 continue;
184 if(palindromes>right)
185 return;
186 printf("%d\n",palindromes);
187 }
188 }
189 }
190 }
191 }
192
193
194
195 int main()
196 {
197 freopen("pprime.in","r",stdin);
198 freopen("pprime.out","w",stdout);
199 scanf("%d%d",&left,&right);
200 get_len();
201 switch(min_len){
202 case 1:
203 do_1();
204 break;
205 case 2:
206 do_2();
207 break;
208 case 3:
209 do_3();
210 break;
211 case 4:
212 do_4();
213 break;
214 case 5:
215 do_5();
216 break;
217 case 6:
218 do_6();
219 break;
220 case 7:
221 do_7();
222 break;
223 case 8:
224 do_8();
225 break;
226 }
227 }