1.0：
View Code1.1（加入了字符串和整数构造大数）：
1 struct BigNum{ 2 #define maxlen 10 3 #define memc(a, b) memcpy(a, b, sizeof(b)) 4 #define mem0(a) memset(a, 0, sizeof(a)) 5 typedef __int64 Num[maxlen + 2]; 6 Num num; 7 char s[maxlen + 2]; 8 BigNum operator+(BigNum num2) { 9 BigNum ans; 10 mem0(ans.num); 11 for(int i = 1; i <= maxlen; i++) { 12 ans.num[i] += num[i] + num2.num[i]; 13 ans.num[i + 1] += ans.num[i] / (int)1e9; 14 ans.num[i] %= (int)1e9; 15 } 16 return ans; 17 } 18 BigNum operator*(BigNum num2) { 19 BigNum ans; 20 mem0(ans.num); 21 for(int i = 1; i <= maxlen; i++) { 22 for(int j = 1; j <= maxlen; j++) { 23 if(i + j - 1 <= maxlen) { 24 ans.num[i + j - 1] += num[i] * num2.num[j]; 25 ans.num[i + j] += ans.num[i + j - 1] / (int)1e9; 26 ans.num[i + j - 1] %= (int)1e9; 27 } 28 } 29 } 30 return ans; 31 } 32 void convert() { 33 int len = strlen(s), cnt = 0; 34 for(int i = len - 1; i >= 0; i -= 9) { 35 int p = 0, x = 0, t = 1; 36 while(i - p >= 0 && p < 9) { 37 x += t * (s[i - p] - '0'); 38 p++; 39 t *= 10; 40 } 41 num[++cnt] = x; 42 } 43 } 44 void inp() { 45 mem0(num); 46 scanf("%s", s); 47 convert(); 48 } 49 void outp() { 50 int p = 1; 51 for(int i = maxlen; i >= 1; i--) { 52 if(num[i]) { 53 p = i; 54 break; 55 } 56 } 57 cout<< num[p]; 58 while(--p) { 59 int a[9] = {0}, x = num[p]; 60 for(int i = 0; i < 9; i++) { 61 a[i] = x % 10; 62 x /= 10; 63 } 64 for(int i = 8; i >= 0; i--) { 65 printf("%d", a[i]); 66 } 67 } 68 } 69 void inttostr(int num) { 70 int cnt = 0; 71 while(num) { 72 s[cnt++] = num % 10 + '0'; 73 num /= 10; 74 } 75 s[cnt] = 0; 76 for(int i = 0, j = cnt - 1; i < j; i++, j--) { 77 swap(s[i], s[j]); 78 } 79 } 80 BigNum(char str[]) { 81 strcpy(s, str); 82 mem0(num); 83 convert(); 84 } 85 BigNum(int n) { 86 inttostr(n); 87 mem0(num); 88 convert(); 89 } 90 BigNum(){} 91 };
1.2（加入了整除，取模，减法，现在可以求两个大数的gcd了）
1 struct BigNum{ 2 #define maxlen 10 3 #define memc(a, b) memcpy(a, b, sizeof(b)) 4 #define mem0(a) memset(a, 0, sizeof(a)) 5 typedef LL Num[maxlen + 2]; 6 Num num; 7 char s[maxlen + 2]; 8 int numCmp(BigNum a, BigNum b) { 9 for(int i = maxlen; i; i--) { 10 if(a.num[i] > b.num[i]) return 1; 11 if(a.num[i] < b.num[i]) return -1; 12 } 13 return 0; 14 } 15 bool operator==(BigNum num2) { 16 BigNum tmp; 17 memc(tmp.num, num); 18 return numCmp(tmp, num2) == 0; 19 } 20 bool operator<(BigNum num2) { 21 BigNum tmp; 22 memc(tmp.num, num); 23 return numCmp(tmp, num2) < 0; 24 } 25 bool operator>(BigNum num2) { 26 BigNum tmp; 27 memc(tmp.num, num); 28 return numCmp(tmp, num2) > 0; 29 } 30 BigNum operator+(BigNum num2) { 31 BigNum ans; 32 mem0(ans.num); 33 for(int i = 1; i <= maxlen; i++) { 34 ans.num[i] += num[i] + num2.num[i]; 35 ans.num[i + 1] += ans.num[i] / (int)1e9; 36 ans.num[i] %= (int)1e9; 37 } 38 return ans; 39 } 40 BigNum operator-(BigNum num2) { 41 BigNum ans; 42 mem0(ans.num); 43 for(int i = 1; i <= maxlen; i++) { 44 ans.num[i] += (int)1e9 + num[i] - num2.num[i]; 45 ans.num[i + 1] += ans.num[i] / (int)1e9 - 1; 46 ans.num[i] %= (int)1e9; 47 } 48 return ans; 49 } 50 BigNum operator*(BigNum num2) { 51 BigNum ans; 52 mem0(ans.num); 53 for(int i = 1; i <= maxlen; i++) { 54 for(int j = 1; j <= maxlen; j++) { 55 if(i + j - 1 <= maxlen) { 56 ans.num[i + j - 1] += num[i] * num2.num[j]; 57 ans.num[i + j] += ans.num[i + j - 1] / (int)1e9; 58 ans.num[i + j - 1] %= (int)1e9; 59 } 60 } 61 } 62 return ans; 63 } 64 BigNum operator/(BigNum num2) { 65 BigNum ans; 66 mem0(ans.num); 67 BigNum thisnum; 68 memc(thisnum.num, num); 69 for(int i = maxlen / 2; i; i--) { 70 int l = 0, r = (int)1e9 - 1; 71 while(l < r) { 72 int m = (l + r + 1) >> 1; 73 ans.num[i] = m; 74 BigNum tmp = ans * num2; 75 if(numCmp(tmp, thisnum) <= 0) l = m; 76 else r = m - 1; 77 } 78 ans.num[i] = l; 79 } 80 return ans; 81 } 82 BigNum operator%(BigNum num2) { 83 BigNum ans; 84 memc(ans.num, num); 85 return ans - ans / num2 * num2; 86 } 87 void convert() { 88 int len = strlen(s), cnt = 0; 89 for(int i = len - 1; i >= 0; i -= 9) { 90 int p = 0, x = 0, t = 1; 91 while(i - p >= 0 && p < 9) { 92 x += t * (s[i - p] - '0'); 93 p++; 94 t *= 10; 95 } 96 num[++cnt] = x; 97 } 98 } 99 void inp() { 100 mem0(num); 101 scanf("%s", s); 102 convert(); 103 } 104 void outp() { 105 int p = 1; 106 for(int i = maxlen; i >= 1; i--) { 107 if(num[i]) { 108 p = i; 109 break; 110 } 111 } 112 cout<< num[p]; 113 while(--p) { 114 int a[9] = {0}, x = num[p]; 115 for(int i = 0; i < 9; i++) { 116 a[i] = x % 10; 117 x /= 10; 118 } 119 for(int i = 8; i >= 0; i--) { 120 printf("%d", a[i]); 121 } 122 } 123 } 124 void inttostr(int num) { 125 int cnt = 0; 126 while(num) { 127 s[cnt++] = num % 10 + '0'; 128 num /= 10; 129 } 130 s[cnt] = 0; 131 for(int i = 0, j = cnt - 1; i < j; i++, j--) { 132 swap(s[i], s[j]); 133 } 134 } 135 BigNum(char str[]) { 136 strcpy(s, str); 137 mem0(num); 138 convert(); 139 } 140 BigNum(int n) { 141 inttostr(n); 142 mem0(num); 143 convert(); 144 } 145 BigNum(){} 146 };
1.3（重写，加减乘，整除，取模， 带符号，2014.12.30）
1 struct BigInt { 2 #define MAXN 100 3 #define DIGIT 8 4 #define D_VAL 100000000 5 #define mem0(a) memset(a, 0, sizeof(a)) 6 #define LL long long 7 typedef int NUM[MAXN + 2]; 8 NUM num; 9 bool flag; 10 int cmp(const int a[], const int b[]) const { 11 for(int i = MAXN; i; i--) { 12 if(a[i] != b[i]) return a[i] - b[i]; 13 } 14 return 0; 15 } 16 bool operator < (const BigInt _A) const { 17 return cmp(num, _A.num) < 0; 18 } 19 bool operator <= (const BigInt _A) const { 20 return cmp(num, _A.num) <= 0; 21 } 22 bool operator == (const BigInt _A) const { 23 return cmp(num, _A.num) == 0; 24 } 25 int &operator [] (int x) { 26 return num[x]; 27 } 28 BigInt() {} 29 BigInt(int x) { 30 mem0(num); 31 int c = 1, p = 1, v = 1; 32 if(x < 0) { 33 flag = 1; 34 x = -x; 35 } 36 else flag = 0; 37 while(x) { 38 num[p] += x % 10 * v; 39 if(c < DIGIT) v *= 10; 40 else v = 1; 41 p += c / DIGIT; 42 c = c % DIGIT + 1; 43 x /= 10; 44 } 45 } 46 BigInt(char s[]) { 47 mem0(num); 48 int len = strlen(s), c = 1, p = 1, v = 1, low = 0; 49 if(s[low] == '-') { 50 low++; 51 flag = 1; 52 } 53 else flag = 0; 54 for(int i = len - 1; i >= low; i--) { 55 num[p] += (s[i] - '0') * v; 56 if(c < DIGIT) v *= 10; 57 else v = 1; 58 p += c / DIGIT; 59 c = c % DIGIT + 1; 60 } 61 } 62 BigInt(int a[], int F) { 63 memcpy(num, a, sizeof(num)); 64 flag = F; 65 } 66 67 void add(int a[], const int b[], const int c[]) const { 68 a[1] = 0; 69 for(int i = 1; i <= MAXN; i++) { 70 a[i] += b[i] + c[i]; 71 a[i + 1] = a[i] / D_VAL; 72 a[i] %= D_VAL; 73 } 74 } 75 void sub(int a[], const int b[], const int c[]) const { 76 a[1] = 0; 77 for(int i = 1; i <= MAXN; i++) { 78 a[i] += b[i] - c[i] + D_VAL; 79 a[i + 1] = a[i] / D_VAL - 1; 80 a[i] %= D_VAL; 81 } 82 } 83 void mul(int a[], const int b[], const int c[]) const { 84 LL aa[MAXN + 2]; 85 mem0(aa); 86 int P = MAXN, Q = MAXN; 87 while(b[P] == 0 && P) P--; 88 while(c[Q] == 0 && Q) Q--; 89 for(int i = 1; i <= P; i++) { 90 for(int j = 1; j <= Q; j++) { 91 int pos = i + j - 1; 92 if(pos <= MAXN) { 93 aa[pos] += (LL)b[i] * c[j]; 94 aa[pos + 1] += aa[pos] / D_VAL; 95 aa[pos] %= D_VAL; 96 } 97 } 98 } 99 for(int i = 1; i <= MAXN; i++) a[i] = aa[i]; 100 } 101 int div(const BigInt &a, const BigInt &b) const { 102 int l = 1, r = D_VAL - 1; 103 while(l < r) { 104 int m = (l + r + 1) >> 1; 105 if(BigInt(m) * b <= a) l = m; 106 else r = m - 1; 107 } 108 return l; 109 } 110 void div(BigInt &a, const BigInt &b, const BigInt &c) const { 111 BigInt rest(0); 112 for(int i = MAXN; i; i--) { 113 rest = BigInt(D_VAL) * rest; 114 rest[1] = b.num[i]; 115 if(rest < c) { 116 a[i] = 0; 117 continue; 118 } 119 a[i] = div(rest, c); 120 rest = rest - BigInt(a[i]) * c; 121 } 122 } 123 BigInt operator + (const BigInt &_A) const { 124 BigInt ans; 125 int tmp = flag << 1 | _A.flag; 126 ans.flag = 0; 127 if(tmp == 0) add(ans.num, num, _A.num); 128 if(tmp == 1) { 129 if(cmp(num, _A.num) >= 0) sub(ans.num, num, _A.num); 130 else { 131 sub(ans.num, _A.num, num); 132 ans.flag = 1; 133 } 134 } 135 if(tmp == 2) { 136 if(cmp(num, _A.num) <= 0) sub(ans.num, _A.num, num); 137 else { 138 sub(ans.num, num, _A.num); 139 ans.flag = 1; 140 } 141 } 142 if(tmp == 3) { 143 ans.flag = 1; 144 add(ans.num, num, _A.num); 145 } 146 return ans; 147 } 148 BigInt operator - (const BigInt &_A) const { 149 BigInt tmp = _A; 150 tmp.flag ^= 1; 151 return *this + tmp; 152 } 153 BigInt operator * (const BigInt &_A) const { 154 BigInt ans; 155 ans.flag = flag ^ _A.flag; 156 mul(ans.num, num, _A.num); 157 return ans; 158 } 159 BigInt operator / (const BigInt &_A) const { 160 BigInt ans; 161 int tmp = flag << 1 | _A.flag; 162 ans.flag = 0; 163 if(tmp == 0) div(ans, *this, _A); 164 if(tmp == 1) { 165 div(ans, *this, _A); 166 ans.flag = 1; 167 } 168 if(tmp == 2) { 169 BigInt T = *this, U = _A; 170 T.flag = 0; 171 U.flag = 1; 172 T = T + _A - BigInt(1); 173 return T / U; 174 } 175 if(tmp == 3) { 176 BigInt T = *this, U = _A; 177 T.flag = U.flag = 0; 178 T = T + _A - BigInt(1); 179 return T / U; 180 } 181 return ans; 182 } 183 BigInt operator % (const BigInt &_A) const { 184 return *this - *this / _A * _A; 185 } 186 void OO(int x) const { 187 int a[DIGIT] = {}, i = 0; 188 while(x) { 189 a[i++] = x % 10; 190 x /= 10; 191 } 192 for(int i = DIGIT - 1; i >= 0; i--) { 193 printf("%d", a[i]); 194 } 195 } 196 void outp() const { 197 if(flag) putchar('-'); 198 int pos = MAXN; 199 while(num[pos] == 0 && pos > 1) pos--; 200 cout << num[pos--]; 201 while(pos) OO(num[pos--]); 202 } 203 void printLine() const { 204 outp(); 205 cout << endl; 206 } 207 void printDigitCount() { 208 int p = MAXN, ans = 0; 209 while(num[p] == 0 && p) p--; 210 if(!p) { 211 puts("1"); 212 return; 213 } 214 int tmp = num[p--]; 215 while(tmp) { 216 ans++; 217 tmp /= 10; 218 } 219 ans += DIGIT * p; 220 printf("%d ", ans); 221 } 222 };