码迷,mamicode.com
首页 > 其他好文 > 详细

P1005-矩阵取数游戏

时间:2019-08-24 11:30:58      阅读:110      评论:0      收藏:0      [点我收藏+]

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 }

 

P1005-矩阵取数游戏

标签:int   lin   lse   syn   out   cout   clu   names   com   

原文地址:https://www.cnblogs.com/Asurudo/p/11403875.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!