大数加减法实现

  1 /*
  2  * input: an expression seperated by a ‘-‘ or ‘+‘; for example: a-b, a+b
  3  * ouput: the answer of the input expression
  4 */
  5 #include <stdio.h>
  6 #include <string.h>
  7 
  8 #define MAX_N 205
  9 
 10 char a[MAX_N], b[MAX_N], ans[MAX_N];
 11 char op;
 12 
 13 void reverse(char *s) {
 14     size_t len = strlen(s);
 15     size_t i;
 16 
 17     for (i = 0; i < (len>>1); i++) {
 18         s[i] ^= s[len-i-1];
 19         s[len-i-1] ^= s[i];
 20         s[i] ^= s[len-i-1];
 21     }
 22 }
 23 
 24 void balance(char *a, char *b) {
 25     size_t len_a = strlen(a);
 26     size_t len_b = strlen(b);
 27 
 28     while (len_a < len_b) {
 29         a[len_a++] = ‘0‘;
 30         a[len_a] = ‘\0‘;
 31     }
 32     while (len_b < len_a) {
 33         b[len_b++] = ‘0‘;
 34         b[len_b] = ‘\0‘;
 35     }
 36 }
 37 
 38 void reduce(char *s) {
 39     size_t len = strlen(s);
 40     
 41     while (len > 1 && s[len-1] == ‘0‘) len--;
 42     s[len] = ‘\0‘;
 43 }
 44 
 45 void add(char *res, char *a, char *b) {
 46     reverse(a);
 47     reverse(b);
 48     balance(a, b);
 49 
 50     size_t n = strlen(a);
 51     size_t i;
 52     int c = 0, x;
 53     char s[MAX_N];
 54 
 55     for (i = 0; i < n; i++) {
 56         x = (a[i] - ‘0‘) + (b[i] - ‘0‘) + c;                
 57         c = x / 10;
 58         s[i] = x % 10 + ‘0‘;                
 59     }
 60     if (c > 0) s[i++] = c + ‘0‘;
 61     s[i] = ‘\0‘;
 62 
 63     reduce(a);
 64     reduce(b);
 65     reverse(a);
 66     reverse(b);
 67     reverse(s);
 68 
 69     strcpy(res, s);
 70 }
 71 
 72 void sub(char *res, char *a, char *b) {
 73     if (strlen(a) < strlen(b) || (strlen(a) == strlen(b) && strcmp(a, b) < 0)) {
 74         res[0] = ‘-‘;
 75         sub(res+1, b, a);
 76         return ;
 77     }    
 78 
 79     reverse(a);
 80     reverse(b);
 81     balance(a, b);
 82 
 83     char s[MAX_N];
 84     size_t n = strlen(a);
 85     size_t i;
 86 
 87     int c = 0, x;
 88     for (i = 0; i < n; i++) {
 89         x = a[i] - b[i] - c;
 90         c = 0;
 91         if (x < 0) {
 92             x += 10;
 93             c = 1;        
 94         }    
 95         s[i] = x + ‘0‘;
 96     }
 97     
 98     s[n] = ‘\0‘;
 99     
100     reduce(s);
101     reduce(b);
102     reverse(a);
103     reverse(b);
104     reverse(s);
105     strcpy(res, s);
106 }
107 
108 void input() {
109     char s[MAX_N];
110     scanf("%s", s);
111 
112     size_t len = strlen(s);
113     size_t i;
114     for (i = 0; i < len; i++) 
115         if (s[i] != ‘-‘ && s[i] != ‘+‘) a[i] = s[i];
116         else break;
117     a[i] = ‘\0‘;
118 
119     op = s[i];
120 
121     strcpy(b, s+i+1);
122 }
123 
124 void output()  {
125     if (op == ‘-‘) 
126         sub(ans, a, b);
127     else
128         add(ans, a, b);
129     puts(ans);
130 }
131 
132 int main(void) {
133     int T;
134     scanf("%d", &T);
135     while ( T-->0) {
136         input();    
137         output();
138     }
139 
140     return 0;
141 }