NOIP模拟水题集锦

大蒟蒻又来水题啦

NOIP2016玩具谜题

 1 #include<stdio.h>
 2 #include<string>
 3 #include<iostream>
 4 using namespace std;
 5 int n,m;
 6 struct toy
 7 {
 8     int dir;
 9     string name;
10 };
11 struct order
12 {
13     int dir;
14     int num;
15 };
16 toy toys[100005];
17 order orders[100005];
18 int main()
19 {
20     scanf("%d%d",&n,&m);
21     int i;
22     for(i=1;i<=n;i++)
23     {
24         scanf("%d",&toys[i].dir);
25         cin>>toys[i].name;
26     }
27     for(i=1;i<=m;i++)
28         scanf("%d%d",&orders[i].dir,&orders[i].num);
29     int now = 1;
30     for(i=1;i<=m;i++)//0 in 1 out 0 left 1 right
31     {
32         if(orders[i].dir)//right
33         {
34             if(toys[now].dir)//out
35             {
36                 now-= orders[i].num;
37                 if(now<=0)now+=n;
38             }
39             else//in
40             {
41                 now+=orders[i].num;
42                 if(now>n) now-=n;
43             }
44         }
45         else//left
46         {
47             if(toys[now].dir)
48             {
49                 now+=orders[i].num;
50                 if(now>n) now-=n;
51             }
52             else
53             {
54                 now-= orders[i].num;
55                 if(now<=0)now+=n;
56             }
57         }
58     }
59     cout<<toys[now].name;
60     return 0;
61 }
View Code

NOIP2015神奇的幻方

 1 #include<stdio.h>
 2 int hf[40][40],n,n2;
 3 int main()
 4 {
 5     scanf("%d",&n); n2 = n*n;
 6     hf[1][(n>>1)+1] = 1;
 7     int prex = 1,prey = (n>>1)+1;
 8     int nowx,nowy;
 9     int i,j;
10     for(i=2;i<=n2;i++)
11     {
12         if(prex==1&&prey!=n)
13         {nowx=n;nowy=prey+1;}
14         else if(prey==n&&prex!=1)
15         {nowy=1;nowx=prex-1;}
16         else if(prex==1&&prey==n)
17         {nowx=2;nowy=n;}
18         else if(prex!=1&&prey!=n)
19         {
20             if(!hf[prex-1][prey+1])
21             {nowx=prex-1;nowy=prey+1;}
22             else{nowx=prex+1;nowy=prey;}
23         }
24         hf[nowx][nowy] = i;
25         prex = nowx;prey = nowy;
26     }
27     for(i=1;i<=n;i++)
28     {
29         for(j=1;j<=n;j++)
30             printf("%d ",hf[i][j]);
31         printf("\n");
32     }
33     return 0;
34 }
View Code

 

posted @ 2017-08-07 17:22  h_zs  阅读(155)  评论(0)  编辑  收藏  举报