Codeforces Beta Round #98 (Div. 2)(A-E)

A

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 105
 #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);
     int ans = ,g=;
     for(i = ; i < k ;i++)
     {
         if(s[i]!=s[i-])
         {
             g = ;
             ans++;
         }
         else g++;
         if(g>)
         {
             g = ;
             ans++;
         }
         //cout<<ans<<" "<<g<<" "<<s[i]<<endl;
     }
     cout<<ans<<endl;
     return ;
 }

B

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 5050
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 int a[N],f[N];
 int main()
 {
     int i,j,n,k;
     cin>>n;
     for(i =;i <= n; i++)cin>>a[i];
     for(i = ;i <= n ;i++)
     {
         f[a[i]] = ;
     }
     int ans=;
     for(i = ; i<=  n; i++)
     if(!f[i]) ans++;cout<<ans<<endl;
     return ;
 }

C

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 100010
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 struct node
 {
     int x,y;
 }p[N];
 bool cmp(node a,node b)
 {
     return a.x<b.x;
 }
 int main()
 {
     int i,j,n;
     cin>>n;
     for(i = ; i <= n ;i++)
     cin>>p[i].x>>p[i].y;
     sort(p+,p+n+,cmp);
     int mm = ,ans=;
     for(i = ; i <= n ;i ++)
     {
         if(p[i].y<mm)
         ans++;
         mm = max(mm,p[i].y);

     }
     cout<<ans<<endl;
     return ;
 }

D

dp  先预处理出来i-j距离回文串所差的步数  然后D出1-k所需最少的步数 记录一下路径 最后输出

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 505
 #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 dp[][],o[][],e[][];
 int p[];
 int dfs(int i,int j)
 {
     int a,b;
     int o1 = ;
     int ii = i,jj = j;
     while(ii<jj)
     {
         if(s[ii]!=s[jj]) o1++;
         ii++,jj--;
     }
     return o1;
 }
 int main()
 {
     int i,j,n,kk,k,g;
     for(i =  ;i <=  ; i++)
         for(j =  ; j <= ; j++)
         dp[i][j] = INF;
     cin>>s;
     kk = strlen(s);
     cin>>k;
     for(i =  ;i < kk ;i++)
         for(j = i ; j < kk ;j++)
         o[i][j] = dfs(i,j);
     for(i = ; i < kk; i++)
         {
             dp[i][] = o[][i];
         }
     for(i =  ;i < kk ;i++)
         for(j =  ; j <= min(i+,k) ; j++)
         for(g =  ; g < i ; g++)
         {
             if(dp[i][j]>dp[g][j-]+o[g+][i])
             {
                 dp[i][j] = dp[g][j-]+o[g+][i];
                 e[i][j] = g;
             }
         }
     int minz = INF,x;
     for(i =  ; i <= k ;i++)
     {
         if(minz>dp[kk-][i])
         {
             minz = dp[kk-][i];
             x = i;
         }
     }
     cout<<minz<<endl;
     if(x==)
     {
         for(i = ;  i < kk/ ; i++)
         printf("%c",s[i]);
         for(i = (kk-)/ ; i >=  ; i--)
         printf("%c",s[i]);
         return ;
     }
     g = ;
     int ki = kk-;
     while(x!=)
     {
         ki = e[ki][x];
         p[g++] = ki;
         x--;
     }
     sort(p,p+g);
     for(i =  ;i <= p[]/ ; i++)
     printf("%c",s[i]);
     if(p[])
     {
         for(i = (p[]-)/ ; i >=  ; i--)
         printf("%c",s[i]);
     }
     for(i =  ; i < g- ;i++)
     {
         printf("+");
         for(j = p[i]+ ;j <= p[i]+(p[i+]-p[i]+)/ ; j++)
         cout<<s[j];
         for(j = p[i]+(p[i+]-p[i])/ ;j >= p[i]+ ; j--)
         cout<<s[j];
     }
     if(p[g-]+<kk)
     printf("+");
     for(i = p[g-]+ ; i <= p[g-]+(kk-p[g-])/ ; i++)
     cout<<s[i];
     for(i = p[g-]+(kk-p[g-]-)/ ; i >= p[g-]+ ; i--)
     cout<<s[i];
     cout<<endl;
     return ;
 }

E

元音-1 辅音+2 开数组sum[i]存前i项和  也就是找最大段(i,j)使sum[j]-sum[i]>=0

这样只存下降的sum数组即可  因为上升的的某个g事不需要的 二分查找第一个小于等于sun[i]的位置 取最大

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 200010
 #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 sum[N];
 struct node
 {
     int a,po;
 }p[N];
 int bfind(int l,int h,int k)
 {
     int m;
     while(l<h)
     {
         m = (l+h)>>;
         if(p[m].a<k)
         h = m;
         else if(p[m].a>k)
         l = m+;
         else return p[m].po;
     }
     return p[l].po;
 }
 int judge(char c)
 {
     if(c=='a'||c=='e'||c=='i'||c=='o'||c=='u')
     return ;
     if(c=='A'||c=='E'||c=='I'||c=='O'||c=='U')
     return ;
     return ;
 }
 int main()
 {
     int i,j,k;
     cin>>s;
     k = strlen(s);
     int g = ;
     for(i = ; i < k ;i++)
     {
         if(i==)
         {
             if(judge(s[i]))
             sum[] = -;
             else
             sum[] = ;
             p[++g].a = sum[i];
             p[g].po = i;
         }
         else
         {
             if(judge(s[i]))
             sum[i] = sum[i-]-;
             else
             sum[i] = sum[i-]+;
             if(sum[i]<p[g].a)
             {
                 p[++g].a = sum[i];
                 p[g].po = i;
             }
         }
     }
     int ans=,cnt=;
     for(i =  ; i < k ;i++)
     {
         if(sum[i]>=)
         {
             ans = i+;
             cnt = ;
         }
         else
         {
             int o = i-bfind(,g,sum[i]);
             //cout<<o<<" "<<i<<" "<<bfind(1,g,sum[i])<<endl;
             if(o>ans)
             {
                 ans = o;
                 cnt = ;
             }
             else if(o==ans)
             cnt++;
         }
     }
     if(ans)
     cout<<ans<<" "<<cnt<<endl;
     else
     puts("No solution");
     return ;
 }