hdu4951 Multiplication table (乘法表的奥秘)

http://acm.hdu.edu.cn/showproblem.php?pid=4951

2014多校 第八题 1008

2014 Multi-University Training Contest 8

Multiplication table

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 438    Accepted Submission(s): 207


Problem Description
Teacher Mai has a multiplication table in base p.

For example, the following is a multiplication table in base 4:

* 0 1 2 3
0 00 00 00 00
1 00 01 02 03
2 00 02 10 12
3 00 03 12 21


But a naughty kid maps numbers 0..p-1 into another permutation and shuffle the multiplication table.

For example Teacher Mai only can see:

1*1=11 1*3=11 1*2=11 1*0=11
3*1=11 3*3=13 3*2=12 3*0=10
2*1=11 2*3=12 2*2=31 2*0=32
0*1=11 0*3=10 0*2=32 0*0=23


Teacher Mai wants you to recover the multiplication table. Output the permutation number 0..p-1 mapped into.

It's guaranteed the solution is unique.
 

Input
There are multiple test cases, terminated by a line "0".

For each test case, the first line contains one integer p(2<=p<=500).

In following p lines, each line contains 2*p integers.The (2*j+1)-th number x and (2*j+2)-th number y in the i-th line indicates equation i*j=xy in the shuffled multiplication table.

Warning: Large IO!
 

Output
For each case, output one line.

First output "Case #k:", where k is the case number counting from 1. The following are p integers, indicating the permutation number 0..p-1 mapped into.
 

Sample Input
4 2 3 1 1 3 2 1 0 1 1 1 1 1 1 1 1 3 2 1 1 3 1 1 2 1 0 1 1 1 2 1 3 0
 

Sample Output
Case #1: 1 3 2 0
 

Source
 

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大意:给出一个p进制数的乘法表,第i行第j组(一组有两个数)表示i*j结果的高位和低位,例如样例中:

4
2 3 1 1 3 2 1 0
1 1 1 1 1 1 1 1
3 2 1 1 3 1 1 2
1 0 1 1 1 2 1 3
0
第二行的“2 3”表示0*0的结果为23,后面的“1 1”表示0*1的结果为11,依此类推。

这个乘法表不正确,是因为有熊孩子把数字打乱了,样例中,0被换成了1, 1被换成了3, 2被换成了2, 3被换成了0,第一行的0*0等于23其实是3*3=21(四进制)。
求各个数的映射。

题解:探寻乘法表的规律。
数字太大,不能暴搜,而且乘法表是有规律的,我们可以根据数字出现的种类数来判断数字。
样例中有一行全是1,我们可以想到1其实就是0,。
最后一行的高位全是1,我们根据10进制的乘法表来观察,什么时候高位会不变呢?只有0*x和1*x高位是不变的,2*x的时候高位就有0或者1了。所以这个高位全是1的行代表的数字是1。
然后探寻普遍的规律:高位出现了多少种数字,这一位就是多少。(0、1高位都只出现1位数字,需特殊判断两个低位是否相等,相等说明是0,不等是1)

(这题有点碉,我是没想到这么好的判断方法,当时暴搜了起来,果然不得力)

代码(一秒多过,加了读入输出优化的话可以到三百多毫秒):
 1 //#pragma comment(linker, "/STACK:102400000,102400000")
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<iostream>
 5 #include<cstring>
 6 #include<algorithm>
 7 #include<cmath>
 8 #include<map>
 9 #include<set>
10 #include<stack>
11 #include<queue>
12 using namespace std;
13 #define ll long long
14 #define usint unsigned int
15 #define mz(array) memset(array, 0, sizeof(array))
16 #define minf(array) memset(array, 0x3f, sizeof(array))
17 #define REP(i,n) for(i=0;i<(n);i++)
18 #define FOR(i,x,n) for(i=(x);i<=(n);i++)
19 #define RD(x) scanf("%d",&x)
20 #define RD2(x,y) scanf("%d%d",&x,&y)
21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
22 #define WN(x) printf("%d\n",x);
23 #define RE  freopen("D.in","r",stdin)
24 #define WE  freopen("1biao.out","w",stdout)
25 #define mp make_pair
26 
27 int a[500][500][2];
28 int c[500];
29 int d[500];
30 int p;
31 
32 void farm() {
33     int i,j;
34     memset(c,-1,sizeof(c));
35 
36     for(i=0; i<p; i++) {
37         memset(d,0,sizeof(d));
38         int cnt=0;
39         for(j=0; j<p; j++)
40             if(d[ a[i][j][1] ]==0) {
41                 d[ a[i][j][1] ]=1;
42                 cnt++;
43             }
44         if(cnt==1) cnt= a[i][0][0]==a[i][1][0]?0:1;
45         //b[i]=cnt;
46         c[cnt]=i;
47     }
48 
49     for(i=0; i<p; i++) {
50         printf(" %d",c[i]);
51     }
52     puts("");
53 }
54 
55 int main() {
56     int cas=1, ans;
57     int i,j;
58     while(scanf("%d",&p)!=EOF) {
59         if(p==0)break;
60         for(i=0; i<p; i++) {
61             for(j=0; j<p; j++) {
62                 //scanf("%d%d",&a[i][j][1],&a[i][j][0]);
63                 scanf("%d%d",&a[i][j][1],&a[i][j][0]);
64             }
65         }
66         printf("Case #%d:",cas++);
67         farm();
68     }
69     return 0;
70 }
View Code

 

 
posted @ 2014-08-15 12:41  带鱼Yuiffy  阅读(799)  评论(0编辑  收藏  举报