头条笔试

updata:

2018-5-12

免费馅饼

 1 #include<bits/stdc++.h>
 2 #define cl(a,b) memset(a,b,sizeof(a))
 3 #define debug(a) cerr<<#a<<"=="<<a<<endl
 4 using namespace std;
 5 typedef long long ll;
 6 typedef pair<int,int> 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<n;i++)
18     {
19         for(int j=0;j<m;j++)
20         {
21             scanf("%d",&mp[i][j]);
22         }
23     }
24     for(int i=0;i<n;i++)
25     {
26         dp[i][0][0]=mp[i][0];
27         dp[i][0][1]=mp[i][0]*-1;
28     }
29     for(int j=1;j<=m;j++)
30     {
31         dp[0][j][0]=max(dp[1][j-1][0],dp[0][j-1][0])+mp[0][j];
32         dp[0][j][1]=max(max(dp[1][j-1][0],dp[0][j-1][0])+mp[0][j]*-1,
33                         max(dp[1][j-1][1],dp[0][j-1][1])+mp[0][j]);
34 
35         dp[n-1][j][0]=max(dp[n-1][j-1][0],dp[n-2][j-1][0])+mp[n-1][j];
36         dp[n-1][j][1]=max(max(dp[n-1][j-1][0],dp[n-2][j-1][0])+mp[n-1][j]*-1,
37                           max(dp[n-1][j-1][1],dp[n-2][j-1][1])+mp[n-1][j]);
38 
39         for(int i=1;i<n-1;i++)
40         {
41             dp[i][j][0]=max(dp[i][j-1][0],max(dp[i-1][j-1][0],dp[i+1][j-1][0]))+mp[i][j];
42 
43             dp[i][j][1]=max(max(dp[i][j-1][0],max(dp[i-1][j-1][0],dp[i+1][j-1][0]))+mp[i][j]*-1,
44                             max(dp[i][j-1][1],max(dp[i-1][j-1][1],dp[i+1][j-1][1]))+mp[i][j]);
45 
46         }
47     }
48     int ans=-1;
49     for(int i=0;i<n;i++)
50     {
51         for(int j=0;j<=m;j++)
52         {
53 //            printf("%d ",dp[i][j][1]);
54             ans=max(ans,max(dp[i][j][0],dp[i][j][1]));
55         }
56 //        printf("\n");
57     }
58     printf("%d\n",ans);
59     return 0;
60 }/*
61 
62 4 3
63 1 -4 10
64 3 -2 -1
65 2 -1 0
66 0 5 -2
67 
68 */

 

 

----------------------------------------------------------------

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<bits/stdc++.h>
 2 #define cl(a,b) memset(a,b,sizeof(a))
 3 #define debug(a) cerr<<#a<<"=="<<a<<endl
 4 using namespace std;
 5 typedef long long ll;
 6 typedef pair<int,int> pii;
 7 
 8 const int maxn=1e6+1000;
 9 
10 set<int>st;
11 
12 int main()
13 {
14     int n,k,x;
15     scanf("%d%d",&n,&k);
16     for(int i=0;i<n;i++)
17     {
18         scanf("%d",&x);
19         st.insert(x);
20     }
21     int ans=0;
22     for(auto i:st)
23     {
24         if(st.find(i+k)!=st.end()) ans++;
25     }
26     printf("%d\n",ans);
27     return 0;
28 }/*
29 
30 6 2
31 1 5 3 3 4 2
32 
33 */

 

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 = 1
14     for i in st :
15         if first != 1:
16             line1 += '..'
17             line2 += '..'
18             line3 += '..'
19             line4 += '..'
20             line5 += '..'
21         first = 0
22         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 line1
83     print line2
84     print line3
85     print line4
86     print line5
87 
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<bits/stdc++.h>
 2 #define cl(a,b) memset(a,b,sizeof(a))
 3 #define debug(a) cerr<<#a<<"=="<<a<<endl
 4 using namespace std;
 5 typedef long long ll;
 6 typedef pair<int,int> 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<n;i++)
34     {
35         scanf("%d",&x);
36         high[x]=true;
37         mx=max(mx,x);
38     }
39     cl(dp,0x3f);
40     check();
41     for(int i=(mx<<1);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 2
54 1 3 6
55 
56 */

 

posted @ 2018-03-24 22:55  良将ℓ  阅读(275)  评论(0)  编辑  收藏  举报