题目大意：给一个表达式，判断两个运算数和结果是否超过int类型的最大值。
可以用double类型进行计算。
1 #include <cstdio> 2 #include <climits> 3 4 int main() 5 { 6 #ifdef LOCAL 7 freopen("in", "r", stdin); 8 #endif 9 char str1[300], str2[300], op; 10 double a, b; 11 while (scanf("%s %c %s", str1, &op, str2) != EOF) 12 { 13 printf("%s %c %s ", str1, op, str2); 14 double a, b, c; 15 sscanf(str1, "%lf", &a); 16 sscanf(str2, "%lf", &b); 17 if (op == '+') c = a + b; 18 else if (op == '*') c = a * b; 19 if (a > INT_MAX) printf("first number too big "); 20 if (b > INT_MAX) printf("second number too big "); 21 if (c > INT_MAX) printf("result too big "); 22 } 23 return 0; 24 }
开始是用bignum类做的，可是一直WA，也不知道怎么回事，先贴上代码，以后在研究吧。
1 #include <cstdio> 2 #include <climits> 3 #include <iostream> 4 #include <cstring> 5 #include <algorithm> 6 using namespace std; 7 8 const int maxn = 300; 9 10 struct bign 11 { 12 int len, s[maxn]; 13 14 bign() 15 { 16 memset(s, 0, sizeof(s)); 17 len = 1; 18 } 19 20 bign(int num) 21 { 22 *this = num; 23 } 24 25 bign(const char * num) 26 { 27 *this = num; 28 } 29 30 bign operator = (int num) 31 { 32 char s[maxn]; 33 sprintf(s, "%d", num); 34 *this = s; 35 return *this; 36 } 37 38 bign operator = (const char * num) 39 { 40 len = strlen(num); 41 for(int i = 0; i < len; i++) 42 s[i] = num[len-1-i] - '0'; 43 return *this; 44 } 45 46 bign operator + (const bign& b) const 47 { 48 bign c; 49 c.len = 0; 50 int g = 0; 51 for(int i = 0; g || i < max(len, b.len); i++) 52 { 53 int x = g; 54 if(i < len) x += s[i]; 55 if(i < b.len) x += b.s[i]; 56 c.s[c.len++] = x % 10; 57 g = x / 10; 58 } 59 return c; 60 } 61 62 void clean() 63 { 64 while(len > 1 && !s[len-1]) len--; 65 } 66 67 bign operator - (const bign& b) 68 { 69 bign c; 70 c.len = 0; 71 int g = 0; 72 for(int i = 0; i < len; i++) 73 { 74 int x = s[i] - g; 75 if(i < b.len) x -= b.s[i]; 76 if(x >= 0) g = 0; 77 else 78 { 79 g = 1; 80 x += 10; 81 } 82 c.s[c.len++] = x; 83 } 84 c.clean(); 85 return c; 86 } 87 88 bign operator * (const bign& b) 89 { 90 bign c; 91 c.len = len + b.len; 92 for(int i = 0; i < len; i++) 93 for(int j = 0; j < b.len; j++) 94 c.s[i+j] += s[i] * b.s[j]; 95 for(int i = 0; i < c.len-1; i++) 96 { 97 c.s[i+1] += c.s[i] / 10; 98 c.s[i] %= 10; 99 } 100 c.clean(); 101 return c; 102 } 103 104 bign operator += (const bign& b) 105 { 106 *this = *this + b; 107 return *this; 108 } 109 110 bool operator < (const bign& b) const 111 { 112 if(len != b.len) return len < b.len; 113 for(int i = len-1; i >= 0; i--) 114 if(s[i] != b.s[i]) return s[i] < b.s[i]; 115 return false; 116 } 117 118 bool operator > (const bign& b) const 119 { 120 return b < *this; 121 } 122 123 bool operator <= (const bign& b) const 124 { 125 return !(b < *this); 126 } 127 128 bool operator >= (const bign& b) const 129 { 130 return !(*this < b); 131 } 132 133 bool operator == (const bign& b) const 134 { 135 return !(b < *this) && !(*this < b); 136 } 137 138 bool operator != (const bign& b) const 139 { 140 return (b < *this) || (b > *this); 141 } 142 143 string str() const 144 { 145 string res = ""; 146 for(int i = 0; i < len; i++) 147 res = (char)(s[i] + '0') + res; 148 if(res == "") res = "0"; 149 return res; 150 } 151 }; 152 153 istream& operator >> (istream& in, bign& x) 154 { 155 string s; 156 in >> s; 157 x = s.c_str(); 158 return in; 159 } 160 161 ostream& operator << (ostream& out, const bign& x) 162 { 163 out << x.str(); 164 return out; 165 } 166 167 int main() 168 { 169 #ifdef LOCAL 170 freopen("in", "r", stdin); 171 #endif 172 char op; 173 bign a, b, c; 174 while (cin >> a >> op >> b) 175 { 176 if (op == '+') c = a + b; 177 else if (op == '*') c = a * b; 178 cout << a << " " << op << " " << b << endl; 179 if (a > INT_MAX) printf("first number too big "); 180 if (b > INT_MAX) printf("second number too big "); 181 if (c > INT_MAX) printf("result too big "); 182 } 183 return 0; 184 } 185 186