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2014
02-14

hdu 2721 Persistent Bits[解题报告]C++

Persistent Bits

问题描述 :

WhatNext Software creates sequence generators that they hope will produce fairly random sequences of 16-bit unsigned integers in the range 0�65535. In general a sequence is specified by integers A, B, C, and S, where 1 ≤ A < 32768, 0 ≤ B < 65536, 2 ≤ C < 65536, and 0 ≤ S < C. S is the first element (the seed) of the sequence, and each later element is generated from the previous element. If X is an element of the sequence, then the next element is

(A * X + B) % C

where ‘%’ is the remainder or modulus operation. Although every element of the sequence will be a 16-bit unsigned integer less than 65536, the intermediate result A * X + B may be larger, so calculations should be done with a 32-bit int rather than a 16-bit short to ensure accurate results.

Some values of the parameters produce better sequences than others. The most embarrassing sequences to WhatNext Software are ones that never change one or more bits. A bit that never changes throughout the sequence is persistent. Ideally, a sequence will have no persistent bits. Your job is to test a sequence and determine which bits are persistent.

For example, a particularly bad choice is A = 2, B = 5, C = 18, and S = 3. It produces the sequence 3, (2*3+5)%18 = 11, (2*11+5)%18 = 9, (2*9+5)%18 = 5, (2*5+5)%18 = 15, (2*15+5)%18 = 17, then (2*17+5)%18 = 3 again, and we’re back at the beginning. So the sequence repeats the the same six values over and over:
Decimal  16-Bit Binary
3  0000000000000011
11  0000000000001011
9  0000000000001001
5  0000000000000101
15  0000000000001111
17  0000000000010001
overall  00000000000????1

The last line of the table indicates which bit positions are always 0, always 1, or take on both values in the sequence. Note that 12 of the 16 bits are persistent. (Good random sequences will have no persistent bits, but the converse is not necessarily true. For example, the sequence defined by A = 1, B = 1, C = 64000, and S = 0 has no persistent bits, but it’s also not random: it just counts from 0 to 63999 before repeating.) Note that a sequence does not need to return to the seed: with A = 2, B = 0, C = 16, and S = 2, the sequence goes 2, 4, 8, 0, 0, 0, ….

输入:

There are from one to sixteen datasets followed by a line containing only 0. Each dataset is a line containing decimal integer values for A, B, C, and S, separated by single blanks.

输出:

There are from one to sixteen datasets followed by a line containing only 0. Each dataset is a line containing decimal integer values for A, B, C, and S, separated by single blanks.

样例输入:

2 5 18 3
1 1 64000 0
2 0 16 2
256 85 32768 21845
1 4097 32776 248
0

样例输出:

00000000000????1
????????????????
000000000000???0
0101010101010101
0???000011111???

地址:http://acm.hdu.edu.cn/showproblem.php?pid=2721

题意:给出a、b、c、s。s是初值,每次变化有s = (a*s+b)%c。如此直到重复。这些数都写成16比特的,如果某位在所有数都是0则输出0,是1则输出1,如果都有可能输出问号。直接暴搞就可以。。。

代码:

# include <stdio.h>
 # include <string.h>
 
 
 int vis[70000] ;
 
 
 int main ()
 {
     int a, b, c, s, i, bit ;
     char ch[20] ;
     
     while (~scanf ("%d", &a) && a)
     {
         scanf ("%d%d%d", &b, &c, &s) ;
         memset (vis, 0, sizeof(vis)) ;
         for (i = 0 ; i < 16 ; i++)
             ch[i] = ((s>>(15-i)) & 1) + '0' ;
         while (!vis[s])
         {
             vis[s] = 1 ;
             for (i = 0 ; i < 16 ; i++)
             {
                 bit = ((s>>(15-i)) & 1) ;
                 if (ch[i]-'0' != bit) ch[i] = '?' ;
             }
             s = (((a*s)%c) + (b%c))%c ;
         }
         for (i = 0 ; i < 16 ; i++)
             printf ("%c", ch[i]) ;
         printf ("\n") ;
     }
     return 0 ;
 }

 

解题转自:http://www.cnblogs.com/lzsz1212/p/3304702.html