2019.5.3备战省赛组队训练赛第十八场

传送门

比赛状况:

题解:https://www.cnblogs.com/QLU-ACM/p/10816527.html

A : 

打表找规律(手动艾特mxl)

a1=3,a2=5

a3=7,a4=9

.................

ai=2*i+1,a(i+1)=2*(i+1)+1 (i为奇数)

假设 n 在 ai 中,如果 i 为奇数,输出 odd,反之,输出 even;
解析
 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define INF 0x3f3f3f3f3f3f3f3f
 5  
 6 ll n;
 7  
 8 int main()
 9 {
10     int test;
11     scanf("%d",&test);
12     for(int kase=1;kase <= test;++kase)
13     {
14         scanf("%lld",&n);
15         double d=sqrt(4+4*n);
16         ll x=(d-2)/2;//求根公式
17         if(x*x+2*x-n >= 0)
18             printf("Case %d: %s
",kase,(x&1) ? "odd":"even");
19         else
20             printf("Case %d: %s
",kase,((x+1)&1) ? "odd":"even");
21     }
22     return 0;
23 }
数学公式
 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define INF 0x3f3f3f3f3f3f3f3f
 5 const int maxn=1e6+6;
 6 ll sum[maxn];
 7 ll n;
 8  
 9 int main()
10 {
11     for(int i=1;i;i++)
12     {
13         sum[i]=3*i+i*(i-1);
14         if(sum[i]>=1000000000)
15         {
16             sum[i+1]=INF;
17             break;
18         }
19     }
20  
21     int test;
22     scanf("%d",&test);
23     for(int kase=1;kase <= test;++kase)
24     {
25         scanf("%lld",&n);
26         for(int i=1;i<=31623;i++)
27         {
28             if(sum[i]>=n&&sum[i-1]<n)
29             {
30                 if(sum[i]&1)//sum[i]为奇数,在奇数项
31                     printf("Case %d: odd
",kase);
32                 else
33                     printf("Case %d: even
",kase);
34                 break;
35             }
36         }
37  
38     }
39     return 0;
40 }
求出sum判断(by mxl)
初始,没往数学公式上推导,想到了二分答案,因为sum[i](前i项和)是升序的;

看了一下数据范围,直接让 l = 0, r = 1e9+1;

wa了两发,仔细想了一下,r 太大时,sum[ mid ]会爆long long 的;

所以,以后碰到这种题,要多往数学公式方面推导,不要动不动就交给代码去实现;

缺乏数学思维,会一直菜下去的;
反思

B :

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int utc[5]={0,-5,0,3,8};
 4 const string area[5]={"","Washington","London","Moscow","Beijing"};
 5  
 6 int nowH;
 7 int nowM;
 8 string nowArea;
 9 int ansH;
10 int ansM;
11 string ansArea;
12 char s[10];
13 int day;
14 map<string ,int >mymap;
15  
16 void Solve()
17 {
18     int diff=mymap[ansArea]-mymap[nowArea];
19     ansH=nowH+diff;
20     string ansDay="Today";
21     if(ansH >= 24)
22     {
23         day=1;
24         ansH -= 24;
25         ansDay="Tomorrow";
26     }
27     else if(ansH < 0)
28     {
29         day=-1;
30         ansH += 24;
31         ansDay="Yesterday";
32     }
33     string am="AM";
34     if(ansH >= 12)
35     {
36         ansH -= 12;
37         am="PM";
38     }
39     ansM=nowM;
40     cout<<ansDay<<" ";
41     printf("%d:%02d ",ansH == 0 ? 12:ansH,ansM);
42     cout<<am<<endl;
43 }
44 int main()
45 {
46 //    freopen("C:/Users/14685/Desktop/stdin&&stdout/contest","r",stdin);
47     for(int i=1;i <= 4;++i)
48         mymap[area[i]]=utc[i];
49     int test;
50     scanf("%d",&test);
51     for(int kase=1;kase <= test;++kase)
52     {
53         day=0;
54         scanf("%d:%d%s",&nowH,&nowM,s);
55         cin>>nowArea>>ansArea;
56         if(s[0] == 'A' && nowH == 12)
57             nowH=0;
58         if(s[0] == 'P' && nowH != 12)
59             nowH += 12;
60         printf("Case %d: ",kase);
61         Solve();
62     }
63     return 0;
64 }
模拟

注意一点:中午12点为PM;

补题:C  H

原文地址:https://www.cnblogs.com/violet-acmer/p/10817737.html