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算法竞赛

2015

幻方....

·1 min
ACM 算法竞赛
c语言上机。。。。 c写的幻方。 1/************************************************************************* 2> File Name: code/class/7.c 3> Author: 111qqz 4> Email: rkz2013@126.com 5> Created Time: 2015年11月11日 星期三 19时31分50秒 6************************************************************************/ 7 8#include<stdio.h> 9#include <string.h> 10 11int n; 12int a[105][105]; 13 13 14 15void swap(int *a,int *b) 16{ 17int tmp; 18tmp = *a; 19*a = *b; 20*b = tmp; 21} 22int fix_x( int x,int k,int n) 23{ 24if (k%2==1) 25{ 26if (x==0) 27return n; 28else return x; 29} 30else 31{ 32if (x==n) 33return n+n; 34else return x; 35} 36} 37int fix_y ( int y,int k,int n) 38{ 39if (k<3) 40{ 41if (y==n+1) 42return 1; 43else return y; 44} 45else 46{ 47if (y==2*n+1) 48return n+1; 49else return y; 50} 51// if (y==n+1) 52// return 1; 53// else return y; 54} 55void print() 56{ 57for ( int i = 1 ; i <= n ; i++) 58{ 59for ( int j = 1 ; j <= n ; j++) 60printf("%d ",a[i][j]); 61 61 62printf("n"); 63} 64 64 65} 66 66 67void OddMagic(int n,int x,int y,int k) //k表示4中状态。。。。 68{ 69 69 70int cur ; 71if (k==1) cur = 1; 72if (k==4) cur = n*n+1; 73if (k==3) cur = n*n*2+1; 74if (k==2) cur = n*n*3+1; 75int cnt = 1; 76while (cnt<=n*n) 77{ 78a[x][y]=cur; 79int prex = x; 80int prey = y; 81cur++; 82cnt++; 83x--; 84y++; 85x = fix_x(x,k,n); 86y = fix_y(y,k,n); 87if (a[x][y]) 88{ 89x = prex+1; 90y = prey; 91} 92 92 93} 94 94 95} 96int main() 97{ 98memset(a,sizeof(a),0); 99scanf("%d",&n); 100if (n%2==1) 101{ 102int x = 1; 103int y = n/2+1; 104OddMagic(n,x,y,1); 105} 106else 107{ 108if (n%4==0) 109{ 110for ( int i = 1,num=1 ; i <= n ; i++) 111for ( int j = 1 ; j <= n ; j++,num++) 112a[i][j]=num; 113 114 115for ( int i = 1 ; i <= n ; i++) 116{ 117for ( int j = 1 ; j <= n ; j++) 118{ 119if (i==j||i+j>=n+1) continue; 120int tmp; 121tmp = a[i][j]; 122a[i][j] = a[n+1-i][n+1-j]; 123a[n+1-i][n+1-j] = tmp; 124} 125} 126} 127else 128{ 129int x = 1; 130int y = n/4+1; 131int hn = n/2; 132 133OddMagic(hn,x,y,1); 134OddMagic(hn,x+hn,y,2); 135OddMagic(hn,x,y+hn,3); 136OddMagic(hn,x+hn,y+hn,4); 137 138int m = n/4; 139for ( int i = 1 ; i <= hn ;i++) 140{ 141for ( int j = 1 ; j <= m ; j++) 142{ 143int tmp; 144if (i==m+1&&j;==m) 145{ 146tmp = a[m+1][m+1]; 147a[m+1][m+1] = a[m+1+hn][m+1]; 148a[m+1+hn][m+1] = tmp; 149continue; 150 151} 152tmp = a[i][j]; 153a[i][j] = a[i+hn][j]; 154a[i+hn][j] = tmp; 155// swap(a[i][j],a[i+n][j]); 156} 157} 158 159for ( int i = 1 ; i <= hn ; i++) 160{ 161for ( int j = n ; j>=n-m+2 ; j--) 162{ 163int tmp; 164tmp = a[i][j]; 165a[i][j] = a[i+hn][j]; 166a[i+hn][j] = tmp; 167} 168} 169 170 171 172 173} 174} 175print(); 176 177}

poj 1106 Transmitters (计算几何，叉积||极角排序)

·2 mins
ACM 算法竞赛
Transmitters **Time Limit:** 1000MS **Memory Limit:** 10000K **Total Submissions:** 4817 **Accepted:** 2576 Description In a wireless network with multiple transmitters sending on the same frequencies, it is often a requirement that signals don’t overlap, or at least that they don’t conflict. One way of accomplishing this is to restrict a transmitter’s coverage area. This problem uses a shielded transmitter that only broadcasts in a semicircle.

hdu 1086 A - You can Solve a Geometry Problem too （线段的规范相交&&非规范相交）

·2 mins
ACM 算法竞赛
A - You can Solve a Geometry Problem too **Time Limit:**1000MS **Memory Limit:**32768KB 64bit IO Format:%I64d & %I64u Submit Status Description Many geometry（几何）problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is very easy, after all we are now attending an exam, not a contest :) Give you N (1<=N<=100) segments（线段）, please output the number of all intersections（交点）. You should count repeatedly if M (M>2) segments intersect at the same point.

codeforces #329 div 2 B. Anton and Lines(几何）

·2 mins
ACM 算法竞赛
B. Anton and Lines time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The teacher gave Anton a large geometry homework, but he didn’t do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = k__i*x + b__i. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between _x_1 < _x_2. In other words, is it true that there are1 ≤ i < j ≤ n and x’, y’, such that:

codeforces #329 div 2 A. 2Char (暴力)

·2 mins
ACM 算法竞赛
A. 2Char time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinct letters. Andrew decided to send an article to the magazine, but as he hasn’t written any article, he just decided to take a random one from magazine 26Char. However, before sending it to the magazine 2Char, he needs to adapt the text to the format of the journal. To do so, he removes some words from the chosen article, in such a way that the remaining text can be written using no more than two distinct letters.