标签:int lin lse syn out cout clu names com
1 #include <bits/stdc++.h> 2 #define _for(i,a,b) for(int i = (a);i < b;i ++) 3 #define _rep(i,a,b) for(int i = (a);i > b;i --) 4 #define INF 0x3f3f3f3f 5 typedef long long ll; 6 using namespace std; 7 inline ll read() 8 { 9 ll ans = 0; 10 char ch = getchar(), last = ‘ ‘; 11 while(!isdigit(ch)) last = ch, ch = getchar(); 12 while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - ‘0‘, ch = getchar(); 13 if(last == ‘-‘) ans = -ans; 14 return ans; 15 } 16 inline void write(ll x) 17 { 18 if(x < 0) x = -x, putchar(‘-‘); 19 if(x >= 10) write(x / 10); 20 putchar(x % 10 + ‘0‘); 21 } 22 const int maxn = 50; 23 class HP 24 { 25 public : int len,s[maxn]; HP() {(*this) = 0;}; 26 HP(int inte) {(*this)=inte;}; HP(const char*str) {(*this)=str;}; 27 friend ostream& operator << (ostream &cout,const HP &x); 28 HP operator = (int inte); HP operator = (const char*str); 29 HP operator * (const HP &b);HP operator + (const HP &b); 30 HP operator - (const HP &b);HP operator / (const HP &b); 31 HP operator % (const HP &b);int Compare(const HP &b); 32 bool operator < (const HP &b); 33 }; 34 35 ostream& operator << (ostream &cout,const HP &x) 36 {for(int i = x.len;i >= 1;i --) cout<<x.s[i];return cout;} 37 38 HP HP::operator = (const char *str) 39 { 40 len = strlen(str); 41 for(int i = 1;i <= len;i ++) s[i] = str[len-i]-‘0‘; 42 return (*this); 43 } 44 45 HP HP::operator = (int inte) 46 { 47 if(inte==0) {len = 1;s[1] = 0;return (*this);}; 48 for(len = 0;inte > 0;) {s[++len] = inte%10;inte /= 10;}; 49 return (*this); 50 } 51 52 HP HP::operator * (const HP&b) 53 { 54 int i,j;HP c;c.len = len+b.len; 55 for(i = 1;i <= c.len;i ++) c.s[i] = 0; 56 for(i = 1;i <=len;i ++) for(j = 1;j <=b.len;j ++) c.s[i+j-1]+=s[i]*b.s[j]; 57 for(i = 1;i < c.len;i ++) {c.s[i+1]+=c.s[i]/10;c.s[i]%=10;} 58 while(c.s[i]) {c.s[i+1]=c.s[i]/10;c.s[i]%=10;i ++;} 59 while(i>1&&!c.s[i]) i--;c.len = i; 60 return c; 61 } 62 63 HP HP::operator+(const HP &b) 64 { 65 int i;HP c;c.s[1] = 0; 66 for(i = 1;i <=len || i<=b.len || c.s[i];i ++) 67 { 68 if(i<=len) c.s[i]+=s[i]; 69 if(i<=b.len) c.s[i]+=b.s[i]; 70 c.s[i+1]=c.s[i]/10;c.s[i]%=10; 71 } 72 c.len = i-1;if(c.len==0) c.len = 1; 73 return c; 74 } 75 76 HP HP::operator-(const HP&b) 77 { 78 int i, j;HP c; 79 for(i = 1,j = 0;i <= len;i ++) 80 { 81 c.s[i] = s[i]-j;if(i<=b.len) c.s[i]-=b.s[i]; 82 if(c.s[i]<0){j = 1;c.s[i]+=10;}else j = 0; 83 } 84 c.len = len;while(c.len>1&&!c.s[c.len]) c.len--; 85 return c; 86 } 87 88 int HP::Compare(const HP &y) 89 { 90 if(len>y.len) return 1; 91 if(len<y.len) return -1; 92 int i = len; 93 while((i>1)&&(s[i]==y.s[i])) i--; 94 return s[i]-y.s[i]; 95 } 96 97 bool HP::operator < (const HP &y) 98 { 99 if(this->Compare(y)>=0) return false; 100 return true; 101 } 102 103 HP HP::operator / (const HP&b) 104 { 105 int i,j;HP d(0),c; 106 for(i = len;i > 0;i --) 107 { 108 if(!(d.len==1 && d.s[1]==0)) 109 {for(j = d.len;j > 0;j --) d.s[j+1]=d.s[j];++d.len;} 110 d.s[1] = s[i]; c.s[i] = 0; 111 while((j = d.Compare(b))>=0) 112 {d=d-b;c.s[i]++;if(j==0) break;} 113 } 114 c.len = len;while((c.len>1)&&(c.s[c.len]==0)) c.len--; 115 return c; 116 } 117 118 HP HP::operator%(const HP&b) 119 { 120 int i,j;HP d(0); 121 for(i = len;i > 0;i --) 122 { 123 if(!(d.len==1 && d.s[1]==0)) 124 {for(j = d.len;j > 0;j --) d.s[j+1]=d.s[j];++d.len;} 125 d.s[1] = s[i]; 126 while((j = d.Compare(b))>=0) {d = d-b;if(j==0)break;} 127 } 128 return d; 129 } 130 131 132 int n,m; 133 HP A[81][81]; 134 HP f[81][81][81]; 135 HP base[81]; 136 137 void Base() 138 { 139 base[0] = 1,base[1] = 2; 140 _for(i,2,81) 141 base[i] = base[i-1]*2; 142 } 143 int main() 144 { 145 ios::sync_with_stdio(false); 146 Base(); 147 n = read(), m = read(); 148 _for(i,1,n+1) 149 _for(j,1,m+1) 150 A[i][j] = read(); 151 152 _for(i,1,n+1) 153 _rep(len,m,0) 154 _for(l,1,m-len+2) 155 { 156 int r = l+len-1; 157 if(f[i][l-1][r]+A[i][l-1]*base[m-r+l-1] 158 < f[i][l][r+1]+A[i][r+1]*base[m-r+l-1]) 159 f[i][l][r] = f[i][l][r+1]+A[i][r+1]*base[m-r+l-1]; 160 else 161 f[i][l][r] = f[i][l-1][r]+A[i][l-1]*base[m-r+l-1]; 162 } 163 164 HP ans; 165 ans = 0; 166 _for(i,1,n+1) 167 { 168 HP tmp; 169 tmp = 0; 170 _for(j,1,m+1) 171 if(tmp < f[i][j][j]+A[i][j]*base[m]) 172 tmp = f[i][j][j]+A[i][j]*base[m]; 173 ans = ans+tmp; 174 } 175 cout << ans << endl; 176 return 0; 177 }
标签:int lin lse syn out cout clu names com
原文地址:https://www.cnblogs.com/Asurudo/p/11403875.html