【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

 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<queue>
 #include<vector>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef vector<int> VI;
 #define fi first
 #define se second
 #define MP make_pair
 #define N   410000
 #define MOD 1000000007
 #define eps 1e-8
 #define pi acos(-1)
 #define oo 110000000000000

 char c[N][],a[N];
 int b[N];

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 void swap(int &x,int &y)
 {
     int t=x;x=y;y=t;
 }

 int main()
 {
     //freopen("1.in","r",stdin);
 //    freopen("1.out","w",stdout);
     ll n;
     int q;
     scanf("%lld%d",&n,&q);
     while(q--)
     {
         ll x,y;
         scanf("%lld%lld",&x,&y);
         ll s1=n%;
         ll s2=x%;
         ll s3=y%;
         ll ans;
         if(s1&&s2&&s3) ans=(x-)/*n+(y+)/;
         if(s1&&s2&&s3==) ans=(n*n+)/+x/*n+y/;
         if(s1&&s2==&&s3) ans=(n*n+)/+x/*n+y/-n/;
         if(s1&&s2==&&s3==) ans=x/*n+y/-n/;
         if(s1==&&s2&&s3) ans=(x-)/*n+(y+)/;
         if(s1==&&s2&&s3==) ans=(n*n+)/+x/*n+y/;
         if(s1==&&s2==&&s3) ans=(n*n+)/+x/*n+y/-n/+;
         if(s1+s2+s3==) ans=x/*n+y/-n/;
         printf("%lld\n",ans);
     }
     return ;
 }