Gray code（hdu5375+异或二进制的规律）
Gray code
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 330 Accepted Submission(s): 183
Problem Description
The reflected binary code, also known as Gray code after Frank Gray, is a binary numeral system where two successive values differ in only onebit (binary digit). The reflected binary code was originally designed to prevent spurious output from electromechanical switches. Today, Gray codes are widely used to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems.
Now , you are given a binary number of length n including ‘0’ , ’1’ and ‘?’(? means that you can use either 0 or 1 to fill this position) and n integers(a1,a2,….,an) . A certain binary number corresponds to a gray code only. If the ith bit of this gray code is 1,you can get the point ai.
Can you tell me how many points you can get at most?
For instance, the binary number “00?0” may be “0000” or “0010”,and the corresponding gray code are “0000” or “0011”.You can choose “0000” getting nothing or “0011” getting the point a3 and a4.
Now , you are given a binary number of length n including ‘0’ , ’1’ and ‘?’(? means that you can use either 0 or 1 to fill this position) and n integers(a1,a2,….,an) . A certain binary number corresponds to a gray code only. If the ith bit of this gray code is 1,you can get the point ai.
Can you tell me how many points you can get at most?
For instance, the binary number “00?0” may be “0000” or “0010”,and the corresponding gray code are “0000” or “0011”.You can choose “0000” getting nothing or “0011” getting the point a3 and a4.
Input
The first line of the input contains the number of test cases T.
Each test case begins with string with ‘0’,’1’ and ‘?’.
The next line contains n (1<=n<=200000) integers (n is the length of the string).
a1 a2 a3 … an (1<=ai<=1000)
Each test case begins with string with ‘0’,’1’ and ‘?’.
The next line contains n (1<=n<=200000) integers (n is the length of the string).
a1 a2 a3 … an (1<=ai<=1000)
Output
For each test case, output “Case #x: ans”, in which x is the case number counted from one,’ans’ is the points you can get at most
Sample Input
2 00?0 1 2 4 8 ???? 1 2 4 8
Sample Output
Case #1: 12 Case #2: 15Hinthttps://en.wikipedia.org/wiki/Gray_code http://baike.baidu.com/view/358724.htm
Source
2015 Multi-University Training Contest 7
格雷码：是由二进制当前位对前以为异或得到的；
eg：二进制 (0)10011
格雷码 11010 即：0^1=1; 1^0=1; 0^0=0; 0^1=1; 1^1=0;
所以：有 (0)x???y 和 (0)x??y <x,y=='0','1','?'>
当x,y不是'?'
当x==y的时候：如果xy之间的?是奇数个那么把x后面的值到y都加起来a[i],(x<i<=y);
那么如果是偶数个呢，那就把和减去min(a[i]),(x<i<=y);
eg: 0????0 那么我们应该让串为：010010 ----->对应格雷码：010111
a[i]= 5 4 2 1 3 6 所以sum=4+2+3+6=15;
当x!=y的时候：如果xy之间的?是偶数个那么把x后面的值到y都加起来a[i],(x<i<=y);
那么如果是奇数个呢，那就把和减去min(a[i]),(x<i<=y);
eg: 0???1 那么我们应该让串为：01101 ----->对应格雷码：01011
a[i]= 5 4 1 2 3 所以sum=4+2+3=9;
等于'?'相当于上述情况，你可以在首位加(0);
转载请注明出处:寻找&星空の孩子
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5375
#include <stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 200005
#define LL long long
int t,len,a[N],ca=1;
char str[N];
int main()
{
// freopen("a.txt","r",stdin);
scanf("%d",&t);
while(ca<=t)
{
scanf("%s",str+1);
len=strlen(str+1);
a[0]=0;
for(int i=1;i<=len;i++)
{
scanf("%d",&a[i]);
}
int flag=0,s1,s2,sa=0;
LL sum=0;
str[0]='0';
char st='0',en;
for(int i=1;i<=len;i++)
{
if(str[i]!='?')
{
en=str[i];
s2=a[i];
if(sa>0)
{
sum+=s2;
if(sa&1 && st!=en) sum-=min(s1,s2);
else if(sa%2==0 && st==en) sum-=min(s1,s2);
}
sa=0;
st=str[i];
if(str[i-1]!='?' && str[i]^str[i-1])
sum+=a[i];
}
else
{
sa++;
sum+=a[i];
if(sa==1) s1=a[i];
else s1=min(s1,a[i]);
}
}
printf("Case #%d: %lld\n",ca++,sum);
}
return 0;
}
##1007. Gray code
格雷码 题解 标准格雷码的性质：二进制a1 a2 ... an，对应的格雷码为a1 （a1 xor a2） ... （an-1 xor an） 题目就可以转为O(n)的dp dp[i][j]表示二进制第i位为j时前i位对应最大分数.
当然dp就不要考虑这么多了：
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;
#define ls 2*i
#define rs 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define ULL unsigned long long
#define N 200005
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define rank rank1
const int mod = 1000000007;
int n;
char str[N];
LL dp[N][2],a[N];
int main()
{
int i,j,k,cas = 1,t;
scanf("%d",&t);
while(t--)
{
scanf("%s",str+1);
int len = strlen(str+1);
for(i = 1; i<=len; i++)
scanf("%d",&a[i]);
MEM(dp,0);
for(i = 1; i<=len; i++)
{
if(i==1)
{
if(str[i]=='?')
{
dp[i][0] = 0;
dp[i][1] = a[i];
}
else if(str[i]=='1')
dp[i][1] = a[i];
else
dp[i][0] = 0;
}
else
{
if(str[i]=='?')
{
if(str[i-1]=='1')
{
dp[i][1]=dp[i-1][1];
dp[i][0]=dp[i-1][1]+a[i];
}
else if(str[i-1]=='0')
{
dp[i][1]=dp[i-1][0]+a[i];
dp[i][0]=dp[i-1][0];
}
else
{
dp[i][1]=max(dp[i-1][1],dp[i-1][0]+a[i]);
dp[i][0]=max(dp[i-1][0],dp[i-1][1]+a[i]);
}
}
else if(str[i]=='1')
{
if(str[i-1]=='1')
dp[i][1]=dp[i-1][1];
else if(str[i-1]=='0')
dp[i][1]=dp[i-1][0]+a[i];
else
dp[i][1]=max(dp[i-1][1],dp[i-1][0]+a[i]);
}
else
{
if(str[i-1]=='1')
dp[i][0]=dp[i-1][1]+a[i];
else if(str[i-1]=='0')
dp[i][0]=dp[i-1][0];
else
dp[i][0]=max(dp[i-1][1]+a[i],dp[i-1][0]);
}
}
}
printf("Case #%d: %I64d\n",cas++,max(dp[len][0],dp[len][1]));
}
return 0;
}