Codeforces Beta Round #96 (Div. 2) （A-E）

写份DIV2的完整题解

A

判断下HQ9有没有出现过

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 112
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 char s[N];
 int main()
 {
     int i,j,k;
     cin>>s;
     k = strlen(s);
     for(i = ; i < k ;i++)
     if(s[i]=='H'||s[i]==''||s[i]=='Q')
     break;
     if(i==k)
     puts("NO");
     else
     puts("YES");
     return ;
 }

B

模拟下就OK

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 112
 #define LL long long
 #define INF 0xfffffff
 #define mod 1000003
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 char s[N];
 int a[];
 int main()
 {
     int i,j,k;
     a['>'] = ;
     a['<'] = ;
     a['+'] = ;
     a['-'] = ;
     a['.'] = ;
     a[','] = ;
     a['['] = ;
     a[']'] = ;
     cin>>s;
     k = strlen(s);
     int ans = ;
     for(i = ;i < k ; i++)
     {
        // cout<<a[s[i]]<<endl;
         ans = (ans*+a[s[i]])%mod;
     }
     cout<<ans<<endl;
     return ;
 }

C

题意有点费解 给定操作 使其数字变为字符 先给你字符 问原先数字为多少 模拟。。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 112
 #define LL long long
 #define INF 0xfffffff
 #define mod 256
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 char s[N];
 int a[];
 int main()
 {
     int i,j,k;
     gets(s);
     k = strlen(s);
     int ans = ;
     int kt = ;
     for(i = ;i < k ; i++)
     {
        int x = s[i];
        int y = kt;
        int g = ;kt=;
        memset(a,,sizeof(a));
        while(y)
        {
            a[g++] = y%;
            y/=;
        }
        for(j = ; j <  ;j++)
        kt+=pow(,-j-)*a[j];
        memset(a,,sizeof(a));
        g = ;
        while(x)
        {
            a[g++] = x%;
            x/=;
        }
        int ans =  ;
        for(j =  ; j <  ;j++)
        ans+=pow(,-j-)*a[j];
        kt = (kt-ans+mod)%mod;
        cout<<kt<<endl;
        kt = s[i];
     }
     return ;
 }

D

题意更是费解 大意:你当前在某一个颜色块中 你有两个指向标  一个是向另一块前进的方向 另一个是在本块的前进方向 0块和边外不能走 问m次后你在哪个块中

模拟。。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 112
 #define LL long long
 #define INF 0xfffffff
 #define mod 256
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 char s[][];
 int a[][][],n,k;
 int judge(int x,int y)
 {
     if(x<||x>=n||y<||y>=k)
     return ;
     if(s[x][y]=='') return ;
     return ;
 }
 int main()
 {
     int i,j,m,g;
     cin>>n>>m;
     for(i = ; i < n ;i++)
         cin>>s[i];
     k = strlen(s[]);
     for(i =  ;i < n ;i++)
     {
         for(j = ; j < k ;j++)
         {
             for(g = j ; g >=  ; g--)
             if(s[i][g]!=s[i][j]) break;
             a[i][j][] = g+;
             for(g = i ; g >= ; g--)
             if(s[g][j]!=s[i][j]) break;
             a[i][j][] = g+;
             for(g = j ; g < k; g++)
             if(s[i][g]!=s[i][j]) break;
             a[i][j][] = g-;
             for(g = i ; g < n ;g++)
             if(s[g][j]!=s[i][j]) break;
             a[i][j][] = g-;
         }
     }
     int d1 = ,d2 = ,x = ,y = ;
     char c = s[][];
     int k1 = ;
     while(k1<=m)
     {
         if(d1==||d1==)
         x = a[x][y][d1];
         else y = a[x][y][d1];
         if(d2==||d2==)
         x = a[x][y][d2];
         else y = a[x][y][d2];
         int tx,ty;
         if(d1==)
         {
             tx = x-;ty = y;
         }
         else if(d1==)
         {
             tx = x;ty = y+;
         }
         else if(d1==)
         {
             tx = x+;ty = y;
         }
         else
         {
             tx = x;ty = y-;
         }
         if(!judge(tx,ty))
         {
             //cout<<",";
             if((d2+)%==d1)
             d2 = (d1+)%;
             else
             {
 
                 d1 = (d1+)%;d2 = (d1-+)%;
             }
         }
         else
         {
             c = s[tx][ty];
             x = tx,y = ty;
         }
         k1++;
         //x = tx
         //cout<<c<<" "<<d1<<" "<<d2<<" "<<k1<<" "<<x<<" "<<y<<endl;
     }
     cout<<c<<endl;
     return ;
 }

E

当时的思路貌似不太对 按照zp说的思路又重写了一遍

dp[i][j][0] 表示在i位置已经改变了j次方向为正的最大移动距离 dp[i][j][1]类似表反方向。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 112
 #define LL long long
 #define INF 0xfffffff
 #define mod 256
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 int dp[N][][];
 char s[N];
 int main()
 {
     int i,j,k;
     cin>>s;
     cin>>k;
     int kk = strlen(s);
     for(i =  ; i < kk ; i++)
         for(j =  ; j <= k ; j++)
         dp[i][j][] = dp[i][j][] = -INF;
     if(s[]=='T')
     {
         for(i =  ; i <= k ; i++)
         {
             if(i%!=)
             dp[][i][] = -;
             else
             dp[][i][] = ;
         }
     }
     else
     {
         for(i =  ; i <= k ; i++)
             if(i%==)
             dp[][i][] = ;
             else
             dp[][i][] = ;
     }
     for(i = ; i < kk ;i++)
     {
         if(s[i]=='T')
         {
             dp[i][][] = dp[i-][][];
             dp[i][][] = dp[i-][][];
         }
         else
         {
             dp[i][][] = dp[i-][][]+;
             dp[i][][] = dp[i-][][]-;
         }
         for(j = ; j <= k ; j++)
         {
             if(s[i]=='T')
             {
                 dp[i][j][] = max(dp[i-][j][],dp[i-][j-][]+);
                 dp[i][j][] = max(dp[i-][j][],dp[i-][j-][]-);
             }
             else
             {
                 dp[i][j][] = max(dp[i-][j][]+,dp[i-][j-][]);
                 dp[i][j][] = max(dp[i-][j][]-,dp[i-][j-][]);
             }
         }
     }
     int ans1 = max(dp[kk-][k][],dp[kk-][k][]);
     for(i =  ; i < kk ; i++)
         for(j =  ; j <= k ; j++)
         dp[i][j][] = dp[i][j][] = -INF;
     if(s[]=='T')
     {
         for(i =  ; i <= k ; i++)
         {
             if(i%!=)
             dp[][i][] = ;
             else
             dp[][i][] = ;
         }
     }
     else
     {
         for(i =  ; i <= k ; i++)
         if(i%==)
         dp[][i][] = -;
         else
         dp[][i][] = ;
     }
     for(i = ; i < kk ;i++)
     {
         if(s[i]=='T')
         {
             dp[i][][] = dp[i-][][];
             dp[i][][] = dp[i-][][];
         }
         else
         {
             dp[i][][] = dp[i-][][]+;
             dp[i][][] = dp[i-][][]-;
         }
         for(j = ; j <= k ; j++)
         {
             if(s[i]=='T')
             {
                 dp[i][j][] = max(dp[i-][j][],dp[i-][j-][]+);
                 dp[i][j][] = max(dp[i-][j][],dp[i-][j-][]-);
             }
             else
             {
                 dp[i][j][] = max(dp[i-][j][]+,dp[i-][j-][]);
                 dp[i][j][] = max(dp[i-][j][]-,dp[i-][j-][]);
             }
         }
     }
     int ans2 = max(dp[kk-][k][],dp[kk-][k][]);
     //cout<<ans1<<" "<<ans2<<endl;
     int ans = max(ans1,ans2);
     cout<<ans<<endl;
     return ;
 }