【CF1027B】Numbers on the Chessboard（数学）

题意：给定一个n*n的矩阵与生成矩阵的方式，多次询问同一个方格内(x,y)的数字是多少

题意：浪费人生的矩阵题，找规律

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<string>
 4 #include<cmath>
 5 #include<iostream>
 6 #include<algorithm>
 7 #include<map>
 8 #include<set>
 9 #include<queue>
10 #include<vector>
11 using namespace std;
12 typedef long long ll;
13 typedef unsigned int uint;
14 typedef unsigned long long ull;
15 typedef pair<int,int> PII;
16 typedef vector<int> VI;
17 #define fi first
18 #define se second 
19 #define MP make_pair
20 #define N   410000
21 #define MOD 1000000007
22 #define eps 1e-8 
23 #define pi acos(-1)
24 #define oo 110000000000000
25 
26 char c[N][2],a[N];
27 int b[N];
28 
29 int read()
30 { 
31    int v=0,f=1;
32    char c=getchar();
33    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
34    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
35    return v*f;
36 }
37 
38 
39 void swap(int &x,int &y)
40 {
41     int t=x;x=y;y=t;
42 }
43 
44 
45 
46 int main()
47 {
48     //freopen("1.in","r",stdin);
49 //    freopen("1.out","w",stdout);
50     ll n;
51     int q;
52     scanf("%lld%d",&n,&q);
53     while(q--)
54     {
55         ll x,y;
56         scanf("%lld%lld",&x,&y);
57         ll s1=n%2;
58         ll s2=x%2;
59         ll s3=y%2;
60         ll ans;
61         if(s1&&s2&&s3) ans=(x-1)/2*n+(y+1)/2;
62         if(s1&&s2&&s3==0) ans=(n*n+1)/2+x/2*n+y/2;
63         if(s1&&s2==0&&s3) ans=(n*n+1)/2+x/2*n+y/2-n/2;
64         if(s1&&s2==0&&s3==0) ans=x/2*n+y/2-n/2;
65         if(s1==0&&s2&&s3) ans=(x-1)/2*n+(y+1)/2;
66         if(s1==0&&s2&&s3==0) ans=(n*n+1)/2+x/2*n+y/2;
67         if(s1==0&&s2==0&&s3) ans=(n*n+1)/2+x/2*n+y/2-n/2+1;
68         if(s1+s2+s3==0) ans=x/2*n+y/2-n/2;
69         printf("%lld\n",ans);  
70     }
71     return 0;
72 }

 

posted on 2018-10-24 15:45  myx12345  阅读(225)  评论(0)    收藏  举报

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