标签：com rectangle 分析 ble ack i++ puts iss only
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 548 Accepted Submission(s): 170
Special Judge
let count be a 2d array filled with 0s
iterate through all 1s in the matrix:
suppose this 1 lies in grid(x,y)
iterate every row r:
if grid(r,y)=1:
++count[min(r,x)][max(r,x)]
if count[min(r,x)][max(r,x)]>1:
claim there is a rectangle satisfying the condition
claim there isn‘t any rectangle satisfying the condition
10000 10000 10000 10000 10000
10000 01000 00100 00010 00001
10000 00100 00001 01000 00010
10000 00010 01000 00001 00100
10000 00001 00010 00100 01000
11000 11000 11000 11000 11000
01000 00100 00010 00001 10000
01000 00010 10000 00100 00001
01000 00001 00100 10000 00010
01000 10000 00001 00010 00100
01100 01100 01100 01100 01100
其实就是某种意义上的+1，+2，。。。
#include <iostream> #include <string> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> #include <queue> #include <deque> #include <map> #define range(i,a,b) for(auto i=a;i<=b;++i) #define LL long long #define itrange(i,a,b) for(auto i=a;i!=b;++i) #define rerange(i,a,b) for(auto i=a;i>=b;--i) #define fill(arr,tmp) memset(arr,tmp,sizeof(arr)) using namespace std; int grid[3005][3005],p=47; void init(){ range(i,0,p)range(j,0,p)range(k,0,p)grid[i*p+j][(j*k+i)%p+k*p]=1; } void solve(){ puts("2000"); range(i,0,1999) { range(j,0,1999){ putchar(grid[i][j]+48); if(!((j+1)%5))putchar(‘ ‘); } putchar(‘\n‘); } } int main() { init(); solve(); return 0; }
面向题解编程后，勉强理解了公式。。。这里直接放标程了。。
#include <stdio.h> int P=47,f[233333],an=0,gg[2333][2333]; int main() { for(int i=0;i<P;i++) { for(int r=0;r<P;++r) { ++an; for(int j=i,k=0;k<P;k++,j=(j+r)%P) f[j*P+k]=an; for(int j=0;j<P*P;++j) if(f[j]==an) gg[i*P+r][j]=1; } } printf("%d\n",1999); for(int i=0;i<1999;++i,puts("")) for(int j=0;j<1999;++j) putchar(gg[i+2][j+2]+48); }
标签：com rectangle 分析 ble ack i++ puts iss only
原文地址：https://www.cnblogs.com/Rhythm-/p/9371123.html
