Simple Arithmetics

http://poj.org/problem?id=1396

挺坑的一道题，调试了一天跪在了格式上，要注意的情况很多，样例太没有代表性了，很容易理解错格式。。

要注意的情况：(1)(加法、减法) 水平线 '----' 的数量为它相邻两侧字符串的最大长度。

                    (2)(乘法) 当第二个数为一位数时，’----'的数量同(1); 否则输出两次’----‘，最上面的’-----‘数量同(1),下面的’----‘数量为最后结果的长度。

                    (3)计算减法的时候，结果要去掉前导0。

                    (4)注意右对齐,每组数据后有一个空行。

样例：

(1)  999            1000          100    (2)  123      1240       (3)      8888
      +11                -1           -98            *4        *20                -8880
       ----                ---           ---           ---          ---                   -----
      1010              999             2          492            0                       8

                                                                    2480

                                                                    -----

                                                                    24800

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <algorithm>
  4 #include <string.h>
  5 const int N=100010;
  6 using namespace std;
  7 char s1[N]= {'0'},s2[N]= {'0'};
  8 char s11[N],s22[N],s33[N];
  9 char str[N];
 10 char s[550][1010];
 11 char sd[550][1010];
 12 int m = 0,len = 0;
 13 void add(char *st1,char *st2,int len)//大数加法
 14 {
 15     int j;
 16     for (j = strlen(st1); j < len; j++)
 17         st1[j] = '0';
 18     st1[j] = '\0';
 19     for (j = strlen(st2); j < len; j++)
 20         st2[j] = '0';
 21     st1[j] = '\0';
 22     int c = 0,t = 0;
 23     for (int i = 0; i < len; i++)
 24     {
 25         int temp = st1[i]-'0'+st2[i]-'0'+c;
 26         str[t++] = temp%10+'0';
 27         c = temp/10;
 28     }
 29     if (c!=0)
 30         str[t++] = c+'0';
 31     str[t]='\0';
 32     reverse(str,str+t);
 33 }
 34 void sub(char *st1,char *st2)//大数减法
 35 {
 36     int j,len=strlen(st1);
 37     for (j = strlen(st2); j < len; j++)
 38         st2[j] = '0';
 39     st1[j] = '\0';
 40     int c = 0,t = 0;
 41     for (int i = 0; st1[i]; i++)
 42     {
 43         int temp;
 44         if(st1[i]-'0'==0)
 45         {
 46             if(c==1)
 47             {
 48                 temp = 9-(st2[i]-'0');
 49                 c = 1;
 50             }
 51             else
 52             {
 53                 if(st2[i]-'0'!=0)
 54                 {
 55                     temp = 10-(st2[i]-'0');
 56                     c = 1;
 57                 }
 58                 else
 59                 {
 60                     temp = 0;
 61                     c = 0;
 62                 }
 63             }
 64         }
 65         else
 66         {
 67             if(st1[i]-'0'-c >= st2[i]-'0')
 68             {
 69                 temp = st1[i]-'0'-c-(st2[i]-'0');
 70                 c = 0;
 71             }
 72             else
 73             {
 74                 temp = st1[i]-'0'+10-c-(st2[i]-'0');
 75                 c = 1;
 76             }
 77 
 78         }
 79         str[t++] = temp%10+'0';
 80     }
 81     str[t]='\0';
 82     reverse(str,str+t);
 83 }
 84 void mul(char* st1, int b)//大数与一位数乘法
 85 {
 86     int c = 0,t = 0;
 87     if (b==0)
 88         strcpy(str,"0");
 89     else
 90     {
 91         for (int i = 0; st1[i]; i++)
 92         {
 93             int temp = (st1[i]-'0')*b+c;
 94             str[t++] = temp%10+'0';
 95             c = temp/10;
 96         }
 97         if (c!=0)
 98             str[t++] = c+'0';
 99         str[t] = '\0';
100         reverse(str,str+t);
101     }
102     strcpy(s[m],str);
103     strcpy(sd[m++],str);
104 }
105 inline void show(int len1,int len2,char* str1,char ch)//格式输出
106 {
107     len = len1 > len2+1?len1:len2+1;
108     int len3 = strlen(str1);
109     int MaxLen = len > len3?len:len3;
110     for (int i = 0; i < MaxLen-len1; i++)
111         printf(" ");
112     puts(s11);
113     for(int i = 0; i < MaxLen-len2-1; i++)
114         printf(" ");
115     putchar(ch);
116     puts(s22);
117     len = len2+1>len3?len2+1:len3;
118     for (int i = 0; i < MaxLen-len; i++)
119         printf(" ");
120     for (int i = 0; i < len; i++)
121         printf("-");
122     puts("");
123     for (int i = 0; i < MaxLen-len3; i++)
124         printf(" ");
125     puts(str1);
126 }
127 int main()
128 {
129     //freopen("data1.in", "r", stdin);
130     //freopen("sad.out", "w", stdout);
131     int T;
132     scanf("%d",&T);
133     while(T--)
134     {
135         m = 0;
136         int p2 = 0,pos = 0;
137         char ss[N],oper;
138         scanf("%s",ss);
139         for (int i = 0; ss[i]; i++)
140         {
141             if (ss[i]=='+'||ss[i]=='-'||ss[i]=='*')
142             {
143                 oper=ss[i];
144                 pos = i;
145                 break;
146             }
147             else
148                 s1[i]=ss[i];
149         }
150         s1[pos]='\0';
151         strcpy(s2,ss+pos+1);
152         p2 = strlen(ss)-pos-1;
153         strcpy(s11,s1);
154         strcpy(s22,s2);
155         reverse(s1,s1+pos);
156         reverse(s2,s2+p2);
157         len = pos>p2?pos:p2;
158         if (oper=='+')
159         {
160             add(s1,s2,len);
161             show(pos,p2,str,'+');
162         }
163         else if(oper=='-')
164         {
165             sub(s1,s2);
166             int gps,pp = 0;
167             for (int i = 0; str[i]; i++)
168             {
169                 if(str[i]!='0')
170                 {
171                     gps = i;
172                     break;
173                 }
174             }
175             for (int i = gps; str[i]; i++)
176             {
177                 str[pp++] = str[i];
178             }
179             str[pp] = '\0';
180             show(pos,p2,str,'-');
181         }
182         else if (oper=='*')
183         {
184             int Mlen = len;
185             mul(s1,s2[0]-'0');
186             strcpy(ss,s[0]);
187             for (int i = 1; s2[i]; i++)
188             {
189                 mul(s1,s2[i]-'0');
190                 int lens = strlen(s[i]);
191                 for (int j = 0; j < i; j++)
192                     s[i][lens++] = '0';
193                 s[i][lens] = '\0';
194                 Mlen=max(strlen(s[i]),strlen(ss));
195                 reverse(s[i],s[i]+strlen(s[i]));
196                 reverse(ss,ss+strlen(ss));
197                 add(s[i],ss,Mlen);
198                 strcpy(ss,str);
199             }
200             Mlen = strlen(ss);
201             if (Mlen==p2)
202                 Mlen=p2+1;
203             if(strlen(s22)==1)
204                 show(pos,p2,ss,'*');
205             else
206             {
207                 for (int i = 0; i < Mlen-pos; i++)
208                     printf(" ");
209                 puts(s11);
210                 for (int i = 0; i < Mlen-p2-1; i++)
211                     printf(" ");
212                 printf("*");
213                 puts(s22);
214                 len = p2+1>strlen(sd[0])?p2+1:strlen(sd[0]);
215                 for (int i = 0; i < Mlen-len; i++)
216                     printf(" ");
217                 for (int i = 0; i < len; i++)
218                     printf("-");
219                 puts("");
220                 for (int i = 0; i < m; i++)
221                 {
222                     int d = Mlen-strlen(sd[i])-i;
223                     for (int j = 0; j < d; j++)
224                         printf(" ");
225                     printf("%s\n",sd[i]);
226                 }
227                 for (int i = 0; i < Mlen-strlen(ss); i++)
228                     printf(" ");
229                 for(int i = 0; i < strlen(ss); i++)
230                     printf("-");
231                 puts("");
232                 for (int i = 0; i < Mlen-strlen(ss); i++)
233                     printf(" ");
234                 puts(ss);
235             }
236         }
237         puts("");
238     }
239     return 0;
240 }
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posted @ 2014-02-16 01:53  N_ll  阅读(471)  评论(0)  编辑  收藏  举报