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 ",ans); 70 } 71 return 0; 72 }