今天学长讲的卡特兰数真的是卡的一批，整个全是高精的题，这时我就使用重载运算符，然后一下午就过去了
首先来看一波水题（也就卡了2小时）
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A. 网格
内存限制：512 MiB 时间限制：1000 ms 标准输入输出
题目描述
这道题是把对角线上方的对角线翻折，然后，自己一看就是C（n+m,n）-C(n+m,n-1),然后就是高精部分，我就不说了;
先看一下我一开始打的代码：
前方高能！
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 #include<cstdlib> 6 #include<ctime> 7 using namespace std; 8 #define maxn 10000 9 #define base 10000 10 //可能重载运算符码量稍大qwq,而且巨难调试 11 struct Bigint 12 { 13 int c[maxn],len,sign; 14 Bigint(){memset(c,0,sizeof(c)),len=1,sign=0;} 15 void Zero() 16 { 17 while(len>1&&c[len]==0)len--; 18 if(len==1&&c[len]==0)sign=0; 19 } 20 void Write(char *s) 21 { 22 int k=1,l=strlen(s); 23 for(int i=l-1;i>=0;i--) 24 { 25 c[len]+=(s[i]-'0')*k; 26 k*=10; 27 if(k==base) 28 k=1,len++; 29 } 30 } 31 void Read() 32 { 33 char s[maxn]={0}; 34 scanf("%s",s); 35 Write(s); 36 } 37 void Print() 38 { 39 if(sign)printf("-"); 40 printf("%d",c[len]); 41 for(int i=len-1;i>=1;i--)printf("%d",c[i]); 42 printf(" "); 43 } 44 bool operator < (const Bigint &a)const 45 { 46 if(len!=a.len)return len<a.len; 47 for(int i=len;i>=1;i--) 48 if(c[i]!=a.c[i])return c[i]<a.c[i]; 49 return 0; 50 } 51 bool operator > (const Bigint &a)const 52 { 53 return a<*this; 54 } 55 Bigint operator = (int a) 56 { 57 char s[100]; 58 sprintf(s,"%d",a); //int ->string 59 Write(s); 60 return *this; 61 } 62 Bigint operator + (const Bigint &a) 63 { 64 Bigint r; 65 r.len=max(len,a.len)+1; 66 for(int i=1;i<=r.len;i++) 67 { 68 r.c[i]+=c[i]+a.c[i]; 69 r.c[i+1]+=r.c[i]/base; 70 r.c[i]%=base; 71 } 72 r.Zero(); 73 return r; 74 } 75 Bigint operator + (const int &a) 76 { 77 Bigint b;b=a; 78 return *this+b; 79 } 80 Bigint operator - (const Bigint &a) 81 { 82 Bigint b,c; 83 b=*this; 84 c=a; 85 if(c>b) 86 { 87 swap(b,c); 88 b.sign=1; 89 } 90 for(int i=1;i<=b.len;i++) 91 { 92 b.c[i]=b.c[i]-c.c[i]; 93 if(b.c[i]<0) 94 { 95 b.c[i]+=base; 96 b.c[i+1]--; 97 } 98 } 99 b.Zero(); 100 return b; 101 } 102 Bigint operator - (const int &a) 103 { 104 Bigint b;b=a;return *this-b; 105 } 106 Bigint operator * (const Bigint &a) 107 { 108 Bigint r; 109 r.len=len+a.len+2; 110 for(int i=1;i<=len;i++) 111 { 112 for(int j=1;j<=a.len;j++) 113 r.c[i+j-1]+=c[i]*a.c[j]; 114 } 115 for(int i=1;i<=r.len;i++) 116 { 117 r.c[i+1]+=r.c[i]/base; 118 r.c[i]%=base; 119 } 120 r.Zero(); 121 return r; 122 } 123 Bigint operator * (const int &a) 124 { 125 Bigint b;b=a; 126 return *this*b; 127 } 128 Bigint operator / (const Bigint &b) 129 { 130 Bigint r,t,a; 131 a=b; 132 r.len=len; 133 for(int i=len;i>=1;i--) 134 { 135 t=t*base+c[i]; 136 int div,ll=0,rr=base; 137 while(ll<=rr) 138 { 139 int mid=(ll+rr)/2; 140 Bigint k=a*mid; 141 if(k>t)rr=mid-1; 142 else 143 { 144 ll=mid+1; 145 div=mid; 146 } 147 } 148 r.c[i]=div; 149 t=t-a*div; 150 } 151 r.Zero(); 152 return r; 153 } 154 Bigint operator / (const int &a) 155 { 156 Bigint b;b=a; 157 return *this/b; 158 } 159 }; 160 int main() 161 { 162 //freopen("cd.txt","r",stdin); 163 Bigint a,b,c; 164 int n,m; 165 scanf("%d%d",&n,&m); 166 a=1,b=1; 167 for(int i=m+2;i<=n+m;i++) 168 a=a*i; 169 if(m+1-n>0) 170 a=a*(m+1-n); 171 for(int i=1;i<=n;i++) 172 b=b*i; 173 c=a/b; 174 c.Print(); 175 //cout<<clock()<<endl; 176 return 0; 177 }
然后，我就直接站一下，我今天一下午顺便集齐的BigInt操作：
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 using namespace std; 6 #define maxn 2000 7 #define base 10000 8 9 struct Bign 10 { 11 int c[maxn],len,sign; 12 //初始化 13 Bign(){memset(c,0,sizeof(c)),len = 1,sign = 0;} 14 //高位清零 15 void Zero() 16 { 17 while(len > 1 && c[len] == 0)len--; 18 if(len == 1 && c[len] == 0)sign = 0; 19 } 20 //压位读入 21 void Write(char *s) 22 { 23 int k = 1,l = strlen(s); 24 for(int i = l - 1;i >= 0;i--) 25 { 26 c[len] += (s[i] - '0') * k; 27 k *= 10; 28 if(k == base) 29 { 30 k = 1; 31 len++; 32 } 33 } 34 } 35 void Read() 36 { 37 char s[maxn] = {0}; 38 scanf("%s",s); 39 Write(s); 40 } 41 //输出 42 void Print() 43 { 44 if(sign)printf("-"); 45 printf("%d",c[len]); 46 for(int i = len - 1;i >= 1;i--)printf("%04d",c[i]); 47 printf(" "); 48 } 49 //重载 = 运算符，将低精赋值给高精 50 Bign operator = (int a) 51 { 52 char s[100]; 53 sprintf(s,"%d",a); 54 Write(s); 55 return *this;//this只能用于成员函数，表示当前对象的地址 56 } 57 //重载 < 运算符 58 bool operator < (const Bign &a)const 59 { 60 if(len != a.len)return len < a.len; 61 for(int i = len;i >= 1;i--) 62 { 63 if(c[i] != a.c[i])return c[i] < a.c[i]; 64 } 65 return 0; 66 } 67 bool operator > (const Bign &a)const 68 { 69 return a < *this; 70 } 71 bool operator <= (const Bign &a)const 72 { 73 return !(a < *this); 74 } 75 bool operator >= (const Bign &a)const 76 { 77 return !(*this < a); 78 } 79 bool operator != (const Bign &a)const 80 { 81 return a < *this || *this < a; 82 } 83 bool operator == (const Bign &a)const 84 { 85 return !(a < *this) && !(*this < a); 86 } 87 bool operator == (const int &a)const 88 { 89 Bign b;b = a; 90 return *this == b; 91 } 92 93 //重载 + 运算符 94 Bign operator + (const Bign &a) 95 { 96 Bign r; 97 r.len = max(len,a.len) + 1; 98 for(int i = 1;i <= r.len;i++) 99 { 100 r.c[i] += c[i] + a.c[i]; 101 r.c[i + 1] += r.c[i] / base; 102 r.c[i] %= base; 103 } 104 r.Zero(); 105 return r; 106 } 107 Bign operator + (const int &a) 108 { 109 Bign b;b = a; 110 return *this + b; 111 } 112 //重载 - 运算符 113 Bign operator - (const Bign &a) 114 { 115 Bign b,c;// b - c 116 b = *this; 117 c = a; 118 if(c > b) 119 { 120 swap(b,c); 121 b.sign = 1; 122 } 123 for(int i = 1;i <= b.len;i++) 124 { 125 b.c[i] -= c.c[i]; 126 if(b.c[i] < 0) 127 { 128 b.c[i] += base; 129 b.c[i + 1]--; 130 } 131 } 132 b.Zero(); 133 return b; 134 } 135 Bign operator - (const int &a) 136 { 137 Bign b;b = a; 138 return *this - b; 139 } 140 //重载 * 运算符 141 Bign operator * (const Bign &a) 142 { 143 Bign r; 144 r.len = len + a.len + 2; 145 for(int i = 1;i <= len;i++) 146 { 147 for(int j = 1;j <= a.len;j++) 148 { 149 r.c[i + j - 1] += c[i] * a.c[j]; 150 } 151 } 152 for(int i = 1;i <= r.len;i++) 153 { 154 r.c[i + 1] += r.c[i] / base; 155 r.c[i] %= base; 156 } 157 r.Zero(); 158 return r; 159 } 160 Bign operator * (const int &a) 161 { 162 Bign b;b = a; 163 return *this * b; 164 } 165 //重载 / 运算符 166 Bign operator / (const Bign &b) 167 { 168 Bign r,t,a; 169 a = b; 170 //if(a == 0)return r; 171 r.len = len; 172 for(int i = len;i >= 1;i--) 173 { 174 t = t * base + c[i]; 175 int div,ll = 0,rr = base; 176 while(ll <= rr) 177 { 178 int mid = (ll + rr) / 2; 179 Bign k = a * mid; 180 if(k > t)rr = mid - 1; 181 else 182 { 183 ll = mid + 1; 184 div = mid; 185 } 186 } 187 r.c[i] = div; 188 t = t - a * div; 189 } 190 r.Zero(); 191 return r; 192 } 193 Bign operator / (const int &a) 194 { 195 Bign b;b = a; 196 return *this / b; 197 } 198 //重载 % 运算符 199 Bign operator % (const Bign &a) 200 { 201 return *this - *this / a * a; 202 } 203 Bign operator % (const int &a) 204 { 205 Bign b;b = a; 206 return *this % b; 207 } 208 }; 209 210 int main() 211 { 212 freopen("Bign.in","r",stdin); 213 freopen("Bign.out","w",stdout); 214 Bign a,b,c,d,e,f,g; 215 a.Read(); 216 b.Read(); 217 c = a + b; 218 c.Print(); 219 d = a - b; 220 d.Print(); 221 e = a * b; 222 e.Print(); 223 f = a / b; 224 f.Print(); 225 g = a % b; 226 g.Print(); 227 return 0; 228 }
然后第二题也是卡特兰数的水题，但是.............我又卡了2个小时
重载运算符，确实很好用，但是不能拿重载之前的低精的复杂度来考虑，稍有不慎，重载运算符的复杂度就能高的惊人！