POJ2941 SDUT2371Homogeneous squares
2 #include<string.h>
3 int main()
4 {
5 long i,j,n,g,s,t,a[1001],b[1001];
6 char str[8001];//這里數組開大一點 第一次RT了 數比較大
7 while(scanf("%ld", &n)&&n)
8 {
9 s = 0;
10 t = 0;
11 int flag = 1;
12 scanf("%d", &a[0]);
13 for(i = 1 ; i < n ; i++)
14 {
15 scanf("%ld",&a[i]);
16 b[i] = a[i]-a[i-1];
17 }
18 for(g = 2 ; g <= n ; g++)
19 {
20 scanf("%d", &a[0]);
21 for(j = 1 ;j < n ; j++)
22 {
23 scanf("%d", &a[j]);
24 if(b[j]!=a[j]-a[j-1])
25 {
26 flag = 0;
27 goto in;
28 }
29 }
30 }
31 in:{
32 for(i = g ; i <= n ;i++)
33 gets(str);
34 }
35 if(flag == 1)
36 printf("homogeneous\n");
37 else
38 printf("not homogeneous\n");
39 }
40 return 0;
41 }
這個題很容易TLE 從網上搜了下題解 找了篇比較好的 自己按照思想敲了一遍 搬這里了 也忘記出處了 就不貼原創了
Homogeneous squares
Time Limit: 1000MS Memory limit: 65536K
題目描述
Assume you have a square of size?n?that is divided into?n×n?positions just as a checkerboard. Two positions?(x1,y1)?and?(x2,y2), where?1 ≤ x1,y1,x2,y2?≤ n, are called "independent" if they occupy different rows and different columns, that is,?x1≠x2?and?y1≠y2. More generally,?n?positions are called independent if they are pairwise independent. It follows that there are?n!?different ways to choose?n?independent positions.
Assume further that a number is written in each position of such an?n×n?square. This square is called "homogeneous" if the sum of the numbers written in?n?independent positions is the same, no matter how the positions are chosen. Write a program to determine if a given square is homogeneous!
輸入
The input contains several test cases.?The first line of each test case contains an integer?n?(1 ≤ n ≤ 1000). Each of the next?n?lines contains?n?numbers, separated by exactly one space character. Each number is an integer from the interval?[-1000000,1000000].?
The last test case is followed by a zero.
輸出
For each test case output whether the specified square is homogeneous or not. Adhere to the format shown in the sample output.
示例輸入
2 1 2 3 4 3 1 3 4 8 6 -2 -3 4 0 0示例輸出
homogeneous not homogeneous ?轉載于:https://www.cnblogs.com/shangyu/archive/2012/03/25/2416277.html
總結
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