POJ 3652 Persistent Bits 简单字符串模拟

 

http://poj.org/problem?id=3652

Persistent Bits

Time Limit: 1000MS		Memory Limit: 65536K

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 integersA, 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. IfX 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 3
1 1 64000 0
2 0 16 2
256 85 32768 21845
1 4097 32776 248
0

Sample Output

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

我的WA代码

/* Author : yan
 * Question : POJ 3652 Persistent Bits
 * Data : Tuesday, December 21 2010
*/
#include<stdio.h>
#define bool _Bool
#define true 1
#define false 0

int bit[16];
int bit_cache[16];
int bit_cnt[16];

int visited[65536];
int seed;
int s;

void Deci_to_Bin(int a)
{
	memset(bit_cache,0,sizeof(bit_cache));
	int _i;
	int cnt=15;
	while(a)
	{
		bit_cache[cnt--]=a%2;
		a/=2;
	}
	//for(_i=0;_i<16;_i++) printf("%d",bit_cache[_i]);
	//printf("/n");
}
bool have_visited(int a)
{
	int _i;
	if(a==s) return true;
	for(_i=0;_i<seed;_i++)
		if(a==visited[_i])
			return true;
	return false;
}
int main()
{
	freopen("input","r",stdin);
	int a,b,c;
	int tmp,i;
	int pos;
	while(scanf("%d",&a)&&a!=0)
	{
		seed=0;
		scanf("%d %d %d",&b,&c,&s);
		visited[seed++]=s;

		if(s==0)
		{
			Deci_to_Bin(c-1);
			for(i=0;i<16;i++)
				if(bit_cache[i]==1)
				{
					pos=i;
					break;
				}
			for(i=0;i<pos;i++) printf("%d",bit_cache[i]);
			for(i=pos;i<16;i++) printf("?");
			printf("/n");
			continue;
		}
		Deci_to_Bin(s);
		for(i=0;i<16;i++) bit[i]=bit_cache[i];

		tmp=s;

		memset(bit_cnt,0,sizeof(bit_cnt));

		while(1)
		{
			tmp=(a*tmp+b)%c;

			if(have_visited(tmp)) break;
			Deci_to_Bin(tmp);
			for(i=0;i<16;i++)
				if(bit[i]!=bit_cache[i]) bit_cnt[i]++;
			visited[seed++]=tmp;
		}
		for(i=0;i<16;i++)
		{
			if(bit_cnt[i]!=0) printf("?");
			else printf("%d",bit[i]);
		}
		printf("/n");
	}
	return 0;
}
 

别人的代码，为了方便阅读，暂时贴过来看看

#include<iostream>
using namespace std;

bool flag[100001];
int wk[2][16];

int main()
{
int A=0, S=0 ,S1=0, B=0, C=0,k=16,j=0,yu=0,time=0;
while( scanf("%d",&A) && A != 0 )
{       
scanf("%d%d%d",&B,&C,&S);
S1=0,k=16,j=0,yu=0,time=0;
memset(flag,false,sizeof(flag));
memset(wk,0,sizeof(wk));
if(S==0&&B==0)
{
printf("0000000000000000/n");
continue;
}   
while(flag[S]!=true)
{
k=15;
time++;
yu=S;
while(yu!=1&&yu!=0&&time==1)
{
wk[0][k--]=yu%2;
yu/=2;
wk[0][k]=1;       
}
S1=(A*S+B)%C;
k=15;
yu=S1;
if(S1==0)
{
if( B == 0 )
{
break;
}
for(j=0;j<16;j++)
wk[1][j]=0;

}
if(yu==1)
{
for(j=0;j<16;j++)
wk[1][j]=0;
wk[1][15]=1;
}

while( yu!=1 && S1 != 0 )
{       
wk[1][k--]=yu%2;
yu/=2;
wk[1][k]=1;
}
flag[S]=true;    
for(j=15;j>=0;j--)
{                 
if(wk[0][j]!=wk[1][j])
{                             
wk[1][j]=100;
}       
}

for(j=0;j<16;j++)
wk[0][j]=wk[1][j];
S=S1;
if(time==63999)
break;
} 
for(j=0;j<16;j++) 
{
if(wk[1][j]==100)
{
cout<<"?";
continue;
}
cout<<wk[1][j];
}
cout<<endl; 
}
return 0;
} 

http://hi.baidu.com/wklove123/blog/item/4123bfcc1883205a0fb34594.html

唉，原来没看清题意，又仔细读了读题目  最后AC！！！

/* Author : yan * Question : POJ 3652 Persistent Bits * Data : Tuesday, December 21 2010 */ #include<stdio.h> #define bool _Bool #define true 1 #define false 0 int bit[16]; int bit_cache[16]; int bit_cnt[16]; bool visited[65536]; int s; void Deci_to_Bin(int a) { memset(bit_cache,0,sizeof(bit_cache)); int _i; int cnt=15; while(a) { bit_cache[cnt--]=a%2; a/=2; } } int main() { //freopen("input","r",stdin); //freopen("output","w",stdout); int a,b,c; int tmp,i; int pos; int cnt; while(scanf("%d",&a)&&a!=0) { scanf("%d %d %d",&b,&c,&s); memset(bit_cnt,0,sizeof(bit_cnt)); memset(visited,false,sizeof(visited)); visited[s]=true; if(a==0 && b==0) { printf("0000000000000000/n"); continue; } Deci_to_Bin(s); for(i=0;i<16;i++) bit[i]=bit_cache[i]; tmp=s; cnt=65535; while(cnt--) { tmp=(a*tmp+b)%c; if(visited[tmp]) break; Deci_to_Bin(tmp); for(i=0;i<16;i++) if(bit[i]!=bit_cache[i]) bit_cnt[i]++; visited[tmp]=true; } for(i=0;i<16;i++) { //printf("/nbit_cnt[%d] = %d/n",i,bit_cnt[i]); if(bit_cnt[i]!=0) printf("?"); else printf("%d",bit[i]); } printf("/n"); } //Deci_to_Bin(10); return 0; }