updata:

2018-5-12

免费馅饼

1 #include
     
       2 #define cl(a,b) memset(a,b,sizeof(a)) 3 #define debug(a) cerr<<#a<<"=="<
      <
       
         pii; 7  8 const int maxn=1e4+10; 9 10 int mp[maxn][maxn];11 int dp[maxn][maxn][2];12 13 int main()14 {15     int n,m;16     scanf("%d%d",&n,&m);17     for(int i=0;i
       
     

 

 

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有几个题没有留

答完才发现是按最高分算 不是最后提交的

早知道就不存备份了

 

A.显然没有找到正解 直接set走一波到80%了

lower_bound竟然才30% 玄学啊...

 

1 #include
     
       2 #define cl(a,b) memset(a,b,sizeof(a)) 3 #define debug(a) cerr<<#a<<"=="<
      <
       
         pii; 7  8 const int maxn=1e6+1000; 9 10 set
        
         st;11 12 int main()13 {14     int n,k,x;15     scanf("%d%d",&n,&k);16     for(int i=0;i
        
       
     

 

B.是一个关于字符串处理的题 反正是有两种操作 问最快处理到当前串的步数

算了下复杂度感觉bfs走一波应该没问题 最后70分也懒得优化了

//也是暴力走一波到70 代码没留

 

C.中缀转后缀这种东西我怎么会自己写

 

1 #coding=utf-8 2  3 import sys 4  5  6 def show(num): 7     line1 = "" 8     line2 = "" 9     line3 = ""10     line4 = ""11     line5 = ""12     st = str(num)13     first = 114     for i in st :15         if first != 1:16             line1 += '..'17             line2 += '..'18             line3 += '..'19             line4 += '..'20             line5 += '..'21         first = 022         if i == '0' :23             line1 += '66666'24             line2 += '6...6'25             line3 += '6...6'26             line4 += '6...6'27             line5 += '66666'28         if i == '1' :29             line1 += '....6'30             line2 += '....6'31             line3 += '....6'32             line4 += '....6'33             line5 += '....6'34         if i == '2' :35             line1 += '66666'36             line2 += '....6'37             line3 += '66666'38             line4 += '6....'39             line5 += '66666'40         if i == '3' :41             line1 += '66666'42             line2 += '....6'43             line3 += '66666'44             line4 += '....6'45             line5 += '66666'46         if i == '4' :47             line1 += '6...6'48             line2 += '6...6'49             line3 += '66666'50             line4 += '....6'51             line5 += '....6'52         if i == '5' :53             line1 += '66666'54             line2 += '6....'55             line3 += '66666'56             line4 += '....6'57             line5 += '66666'58         if i == '6' :59             line1 += '66666'60             line2 += '6....'61             line3 += '66666'62             line4 += '6...6'63             line5 += '66666'64         if i == '7' :65             line1 += '66666'66             line2 += '....6'67             line3 += '....6'68             line4 += '....6'69             line5 += '....6'70         if i == '8' :71             line1 += '66666'72             line2 += '6...6'73             line3 += '66666'74             line4 += '6...6'75             line5 += '66666'76         if i == '9' :77             line1 += '66666'78             line2 += '6...6'79             line3 += '66666'80             line4 += '....6'81             line5 += '66666'82     print line183     print line284     print line385     print line486     print line587 88 if __name__ == "__main__":89     n = int(sys.stdin.readline())90     for i in range(n):91         line = sys.stdin.readline()92         res = eval(line)93         # print(res)94         show(res)

 

D.暴力一波走了30分 其他的肝不动了= =

 

E.这个复杂度高的一批 数据太水了吧

（虽然加了一点玄学优化 可能就给优化掉了？

 

1 #include
     
       2 #define cl(a,b) memset(a,b,sizeof(a)) 3 #define debug(a) cerr<<#a<<"=="<
      <
       
         pii; 7  8 const int maxn=1e6+10; 9 10 int n,k,h,mx;11 bool high[maxn<<1];12 int dp[maxn<<1];13 14 void check()15 {16     dp[0]=0;17     for(int i=0;i<=mx;i++)18     {19         for(int j=i+1;j<=i+h;j++)20         {21             if(high[j]) dp[2*j-i]=min(dp[2*j-i],dp[i]+1);22         }23     }24 25 }26 27 int main()28 {29     int x,ans=-1;30     mx=-1;31     scanf("%d%d%d",&n,&k,&h);32 33     for(int i=0;i
        
         =0;i--)42     {43         if(dp[i]<=k)44         {45             ans=i;46             break;47         }48     }49     printf("%d\n",ans);50     return 0;51 }/*52 53 3 3 254 1 3 655 56 */