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] = ' '; 19 for (j = strlen(st2); j < len; j++) 20 st2[j] = '0'; 21 st1[j] = ' '; 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]=' '; 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] = ' '; 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]=' '; 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] = ' '; 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]=' '; 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] = ' '; 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] = ' '; 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 ",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 }