Description
An accutron shows time with four digits, from 0000 to 2359. Every digit is represented by 3*3 characters, including '|'s, '_'s and blanks. When the LCD screen works well, the digits look like the following:
There are two accutrons at hand. One shows the accurate time, and the other is 15 minutes late. For example, at 8:25am, the first accutron shows '0825', while the second shows '0810'.
Unfortunately, there is something wrong with the two LCD screens, namely some parts of the digits missed. Your task is to decide the accurate time, according to the fragmental digits showed on the two accutrons.
_ _ _ _ _ _ _ _
| | | _| _||_||_ |_ ||_||_|
|_| ||_ _| | _||_| ||_| _|
There are two accutrons at hand. One shows the accurate time, and the other is 15 minutes late. For example, at 8:25am, the first accutron shows '0825', while the second shows '0810'.
Unfortunately, there is something wrong with the two LCD screens, namely some parts of the digits missed. Your task is to decide the accurate time, according to the fragmental digits showed on the two accutrons.
Input
The first line of the input is a single integer t (1 <= t <= 20), the number of test cases. Each case contains three lines, indicating the time on the accurate accutron and the time on the slow accutron, separated by a blank column. (Please refer to the Sample Input.)
Output
For each input, print the accurate time with four digits if it can be ensured, or otherwise the string 'Not Sure'.
Sample Input
2
_ _ _ _ _
| _ _|| _ ||
| _ |_ | | _ |_|
_ _ _ _ _ _
||_ _|| _| ||
| _ |_ | || |_|
Sample Output
Not Sure 0825
题意：一块电子表的液晶屏坏了，根据显示的字符判断前面输入的时间减去15分钟是否等于后面的时间，如果结果唯一输出前面的时间，否则就输出“Not Sure”。
思路：暴力；
具体：把0-9这10个数字转换成一维数组，与输入的作对比。
注意：测试数据“0000 2345”，“ _ _ _ _ _ _ _ ”
| ||_| || | | | ||_ |_
|_||_| ||_| |_| | _| _|。
ac代码：
1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 int s1[10][9]= {{0,1,0,1,0,1,1,1,1},{0,0,0,0,0,1,0,0,1},{0,1,0,0,1,1,1,1,0},{0,1,0,0,1,1,0,1,1},{0,0,0,1,1,1,0,0,1},{0,1,0,1,1,0,0,1,1},{0,1,0,1,1,0,1,1,1},{0,1,0,0,0,1,0,0,1},{0,1,0,1,1,1,1,1,1},{0,1,0,1,1,1,0,1,1}}; 5 int main() 6 { 7 int a,b,c,d,e,f,g,h,i,j,l,l1,n,vis[150],vis1[150],vis2[150],vis3[150],vis4[150],vis5[150],vis6[150],vis7[150]; 8 int x1[150],x2[150],x3[150],x4[150],x5[150],x6[150],x7[150],x8[10000],x9[15000],x[150],a1,a2,a3,a4,a5,a6,a7,a8,a9,a0,b1,b2,b3,b4,b5; 9 int y1[15000],y2[15000],y3[15000],y4[10500]; 10 scanf("%d",&n); 11 getchar(); 12 char s[150][100]; 13 while(n--) 14 { 15 a=0,b=0,c=0,d=0,e=0,f=0,g=0,h=0,a1=0,a2=0,a3=0,a4=0,a5=0,a6=0,a7=0,a8=0,a9=1,a0=0,b1=0,b2=0,b3=0,b4=0,b5=0,l=0,l1=0; 16 for(i=0;i<3;i++) 17 gets(s[i]); 18 for(i=0; i<3; i++) 19 { 20 for(j=0; j<3; j++) 21 { 22 if(s[i][j]==' ') 23 vis[a++]=0; 24 else 25 vis[a++]=1; 26 } 27 for(j=3; j<6; j++) 28 { 29 if(s[i][j]==' ') 30 vis1[b++]=0; 31 else 32 vis1[b++]=1; 33 } 34 for(j=6; j<9; j++) 35 { 36 if(s[i][j]==' ') 37 vis2[c++]=0; 38 else 39 vis2[c++]=1; 40 } 41 for(j=9; j<12; j++) 42 { 43 if(s[i][j]==' ') 44 vis3[d++]=0; 45 else 46 vis3[d++]=1; 47 } 48 for(j=13; j<16; j++) 49 { 50 if(s[i][j]==' ') 51 vis4[e++]=0; 52 else 53 vis4[e++]=1; 54 } 55 for(j=16; j<19; j++) 56 { 57 if(s[i][j]==' ') 58 vis5[f++]=0; 59 else 60 vis5[f++]=1; 61 } 62 for(j=19; j<22; j++) 63 { 64 if(s[i][j]==' ') 65 vis6[g++]=0; 66 else 67 vis6[g++]=1; 68 } 69 for(j=22; j<25; j++) 70 { 71 if(s[i][j]==' ') 72 vis7[h++]=0; 73 else 74 vis7[h++]=1; 75 } 76 } 77 for(i=0; i<10; i++) 78 { 79 a9=0; 80 for(j=0; j<10; j++) 81 { 82 a9++; 83 if(vis[j]>s1[i][j]) 84 { 85 a9=0; 86 break; 87 } 88 if(a9==9) 89 x[a0++]=i; 90 } 91 } 92 for(i=0; i<10; i++) 93 { 94 a9=0; 95 for(j=0; j<10; j++) 96 { 97 a9++; 98 if(vis1[j]>s1[i][j]) 99 { 100 a9=0; 101 break; 102 } 103 if(a9==9) 104 x1[a1++]=i; 105 } 106 } 107 for(i=0; i<10; i++) 108 { 109 a9=0; 110 for(j=0; j<10; j++) 111 { 112 a9++; 113 if(vis2[j]>s1[i][j]) 114 { 115 a9=0; 116 break; 117 } 118 if(a9==9) 119 x2[a2++]=i; 120 } 121 } 122 for(i=0; i<10; i++) 123 { 124 a9=0; 125 for(j=0; j<10; j++) 126 { 127 a9++; 128 if(vis3[j]>s1[i][j]) 129 { 130 a9=0; 131 break; 132 } 133 if(a9==9) 134 x3[a3++]=i; 135 } 136 } 137 for(i=0; i<10; i++) 138 { 139 a9=0; 140 for(j=0; j<10; j++) 141 { 142 a9++; 143 if(vis4[j]>s1[i][j]) 144 { 145 a9=0; 146 break; 147 } 148 if(a9==9) 149 x4[a4++]=i; 150 } 151 } 152 for(i=0; i<10; i++) 153 { 154 a9=0; 155 for(j=0; j<10; j++) 156 { 157 a9++; 158 if(vis5[j]>s1[i][j]) 159 { 160 a9=0; 161 break; 162 } 163 if(a9==9) 164 x5[a5++]=i; 165 } 166 } 167 for(i=0; i<10; i++) 168 { 169 a9=0; 170 for(j=0; j<10; j++) 171 { 172 a9++; 173 if(vis6[j]>s1[i][j]) 174 { 175 a9=0; 176 break; 177 } 178 if(a9==9) 179 x6[a6++]=i; 180 } 181 } 182 for(i=0; i<10; i++) 183 { 184 a9=0; 185 for(j=0; j<10; j++) 186 { 187 a9++; 188 if(vis7[j]>s1[i][j]) 189 { 190 a9=0; 191 break; 192 } 193 if(a9==9) 194 x7[a7++]=i; 195 } 196 } 197 for(i=0; i<a0; i++) 198 { 199 for(j=0; j<a1; j++) 200 if((x[i]*1000+x1[j]*100)<2400) 201 { 202 x8[a8++]=x[i]*1000+x1[j]*100; 203 } 204 } 205 for(i=0; i<a4; i++) 206 { 207 for(j=0; j<a5; j++) 208 if((x4[i]*1000+x5[j]*100)<2400) 209 { 210 x9[b1++]=x4[i]*1000+x5[j]*100; 211 } 212 } 213 for(i=0; i<a2; i++) 214 { 215 for(j=0; j<a3; j++) 216 if((x2[i]*10+x3[j])<60) 217 { 218 y1[b2++]=x2[i]*10+x3[j]; 219 } 220 } 221 for(i=0; i<a6; i++) 222 { 223 for(j=0; j<a7; j++) 224 if((x6[i]*10+x7[j])<60) 225 { 226 y2[b3++]=x6[i]*10+x7[j]; 227 } 228 } 229 for(i=0; i<a8; i++) 230 for(j=0; j<b2; j++) 231 { 232 y3[b4++]=x8[i]+y1[j]; 233 } 234 for(i=0; i<b1; i++) 235 for(j=0; j<b3; j++) 236 { 237 y4[b5++]=x9[i]+y2[j]; 238 } 239 int num=0,sum=0; 240 for(i=0; i<b4; i++) 241 { 242 for(j=0; j<b5; j++) 243 { 244 sum=y3[i]; 245 num=y3[i]%100; 246 if(sum<15) 247 { 248 sum+=2400; 249 } 250 if(num<15) 251 { 252 sum=sum-40; 253 } 254 num=sum-15; 255 if(num==y4[j]) 256 { 257 l++; 258 l1=y3[i]; 259 } 260 } 261 } 262 if(l==1) 263 printf("%04d ",l1); 264 else 265 printf("Not Sure "); 266 } 267 }