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hdu 2721(字符串处理，位运算 暴力)

2024-07-17 15:38:39   218人阅读

Persistent Bits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 201    Accepted Submission(s): 116


Problem Description
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, ....
 


 

Input
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.
 


 

Output
There is one line of output for each data set, each containing 16 characters, either ‘1‘, ‘0‘, or ‘?‘ for each of the 16 bits in order, with the most significant bit first, with ‘1‘ indicating the corresponding bit is always 1, ‘0‘ meaning the corresponding bit is always 0, and ‘?‘ indicating the bit takes on values of both 0 and 1 in the sequence.
 


 

Sample Input
2 5 18 31 1 64000 02 0 16 2256 85 32768 218451 4097 32776 2480
 


 

Sample Output
00000000000????1????????????????000000000000???001010101010101010???000011111???
 


 

Source
Mid-Central USA 2007

 

难点：

如何有效率的将十进制转化为16位的二进制，如何判断是否发生了数字循环出现的现象。

易错点：

数组要开的足够大，否则不能正常输出结果。

题意：

给定一个式子：(A * X + B) % C 转化成二进制要是x是可以变化的，通过x的不断变化，式子也

在不断变化。要是二进制的某一位上，保持不变则还是原始数据，否则是 ？将结果输出。

代码如下:

<span style="font-size:14px;">#include<stdio.h>#include<string.h>int f[101000];//数组要开足够大int main(){	int a,b,c,s,i,bit;	char ch[20];	while(~scanf("%d",&a),a)	{		memset(f,0,sizeof(f));		scanf("%d%d%d",&b,&c,&s);		for(i=0;i<16;i++)		ch[i]=((s>>(15-i))&1)+'0';//将十进制高效的转化为16位二进制		while(!f[s])		{			f[s]=1;			s=((a%c)*(s%c)+(b%c))%c;			for(i=0;i<16;i++)			{				bit=(s>>(15-i))&1;				if(bit+'0'!=ch[i])				ch[i]='?';			}		}		for(i=0;i<16;i++)		{			printf("%c",ch[i]);		}		printf("\n");	}	return 0;}</span>


 


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