Codeforces Round #636 (Div. 3)

A

思路：for循环搞一下，从2开始，i*=2,累加和，当n%sum==0就行了，就说明整除了

 1 //-------------------------------------------------
 2 //Created by HanJinyu
 3 //Created Time :二  4/21 22:34:17 2020
 4 //File Name :636A.cpp
 5 //-------------------------------------------------
 6 
 7 #include <stdio.h>
 8 #include <string.h>
 9 #include <iostream>
10 #include <algorithm>
11 #include <vector>
12 #include <queue>
13 #include <set>
14 #include <map>
15 #include <list>
16 #include <map>
17 #include <string>
18 #include <math.h>
19 #include <stdlib.h>
20 #include <time.h>
21 using namespace std;
22 typedef double db;
23 typedef long long ll;
24 const int maxn = 200005;
25 
26 int main()
27 {
28     //freopen("in.txt","r",stdin);
29     //freopen("out.txt","w",stdout);
30     int T;
31     scanf("%d",&T);
32     while(T--){
33         ll n;
34         scanf("%lld",&n);
35         ll sum=0;
36         for(ll i=1;;i*=2)
37         {
38             sum+=i;
39             if(n%sum==0&&i!=1)
40             {
41                 printf("%lld
",n/sum);
42                 break;
43             }
44         }
45     }
46 
47     return 0;
48 }

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B

思路：看例题的结果解析没啥用，自己找例子，2468，和1379，这俩和相等，你就发现偶数的首尾之和等于奇数的首尾之和，由外两边向内两边每取两个数和都为偶数首尾之和，那么偶数部分正常等差输出即可，奇数分布就把每组19，37，插入动态数组，sort一下输出就好了

 1 //-------------------------------------------------
 2 //Created by HanJinyu
 3 //Created Time :二  4/21 23:17:11 2020
 4 //File Name :636B.cpp
 5 //-------------------------------------------------
 6 
 7 #include <stdio.h>
 8 #include <string.h>
 9 #include <iostream>
10 #include <algorithm>
11 #include <vector>
12 #include <queue>
13 #include <set>
14 #include <map>
15 #include <list>
16 #include <map>
17 #include <string>
18 #include <math.h>
19 #include <stdlib.h>
20 #include <time.h>
21 using namespace std;
22 typedef double db;
23 typedef long long ll;
24 const int maxn = 200005;
25 
26 int main()
27 {
28     //freopen("in.txt","r",stdin);
29     //freopen("out.txt","w",stdout);
30     int t;
31     scanf("%d",&t);
32     while(t--)
33     {
34         int n;
35         scanf("%d",&n);
36         if(n%4==0)
37         {
38             printf("YES
");
39             vector<ll>vv;
40             for(ll i=2;i<=2*(n/2);i+=2)
41                 printf("%lld ",i);
42             for(int i=0;i<n/4;i++)
43             {
44                 ll mm=i*2+1;
45                 ll nn=2+n;
46                 vv.push_back(mm);
47                 vv.push_back(nn-mm);
48             }
49             sort(vv.begin(),vv.end());
50             for(int i=0;i<vv.size();i++)
51                 printf("%lld ",vv[i]);
52             printf("
");
53         } else
54             printf("NO
");
55     }
56 
57     return 0;
58 }

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C

思路：就是在数组里找长度尽可能大的序列+-+-+-+-.../-+-+-+...,使其和最大，那么只要找每个相同符号的一段里面的最大值即可，将其相加，注意下什么时候加值即可

 1 //-------------------------------------------------
 2 //Created by HanJinyu
 3 //Created Time :二  4/21 23:17:11 2020
 4 //File Name :636B.cpp
 5 //-------------------------------------------------
 6 
 7 #include <stdio.h>
 8 #include <string.h>
 9 #include <iostream>
10 #include <algorithm>
11 #include <vector>
12 #include <queue>
13 #include <set>
14 #include <map>
15 #include <list>
16 #include <map>
17 #include <string>
18 #include <math.h>
19 #include <stdlib.h>
20 #include <time.h>
21 using namespace std;
22 typedef double db;
23 typedef long long ll;
24 const int maxn = 200005;
25 const ll xiao=-1e9;
26 int main()
27 {
28     //freopen("in.txt","r",stdin);
29     //freopen("out.txt","w",stdout);
30     int t;
31     scanf("%d",&t);
32     while(t--) {
33         int n;
34         scanf("%d", &n);
35         ll a[maxn];
36         ll zheng=0,fu=xiao,sum=0;
37         for(int i=0;i<n;i++)
38         {
39             scanf("%lld",&a[i]);
40         }
41         a[n]=0;
42         for(int i=0;i<=n-1;i++)
43         {
44             if(a[i]>0&&a[i]>zheng)
45             {
46                 zheng=a[i];
47             }
48             if(a[i]<0&&a[i]>fu)
49             {
50                 fu=a[i];
51             }
52             if(a[i]<0&&a[i+1]>=0)
53             {
54                 sum+=fu;fu=xiao;
55             }
56             else if(a[i]>0&&a[i+1]<=0)
57             {
58                 sum+=zheng;zheng=0;
59             }
60         }
61         if(n==1)
62             printf("%lld
",a[0]);
63         else {
64             printf("%lld
", sum);
65         }
66     }
67 
68     return 0;
69 }

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