哈尔滨理工大学第七届程序设计竞赛（G.Great Atm）
Description
An old story said the evil dragon wasn’t evil at all, only bewitched, and now that the riddles were solved it was proving to be as kind as its new master.
A powerful warrior Atm is going to solve the riddles. First, he should beat the evil wizard. The road from Atm’s castle to wizard’s lab is filled with magic traps. The magic trap will affect Atm’s combat effectiveness.
Atm’s combat effectiveness can be considered as an integer. Effect of magic trap can be considered as mathematical operation. The three kinds of magic traps correspond to three kind of bit operation. (AND, OR and XOR)
Atm can adjust his equipment to change his initial combat effectiveness from 0 to m (include 0 and m). He wants when he arrives the wizard’s lab, his combat effectiveness can be maximum.
Input
There are multiple test cases.
For each test cases:
The first line contains two integers n(1<=n<=10^5) and m(1<=m<=10^9), indicating the number of magic traps and the maximum of initial combat effectiveness.
Each of the next n lines contains a string and an integer, indicating the bit operation. The string will be “AND”, “OR” or “XOR” correspond to AND operation (&), OR operation (|) or XOR operation (^). The integer t(1<=t<=10^9) is second operand in the operation.
Output
For each test cases, a line contains an integer, indicating the maximum combat effectiveness when he arrives the wizard’s lab.
Sample Input
3 10
AND 5
OR 6
XOR 7
Sample Output
1
//哎，自己太菜了，看了隔壁学姐的博客，然后发现，我曹。。。。
//其实这一道题可以用贪心来解释，要从0~m的范围内找一个满足条件的数，不如先从一个数的二进制中想办法，
//二进制中要么是0要么是1，所以从最低位开始往最高位慢慢找.
//每次与操作中要求的操作进行运算，如果一个数的某一位是0经过运算之后为1，那么肯定要把这个位置取0，
//相反则取1。
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<cstdio>
#include<sstream>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include <ctype.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;
const int maxn=110000;
typedef long long ll;
const int INF=0x3f3f3f3f;
struct node
{
char str[15];
int x;
} c[maxn];
int a[32];
void init()
{
a[0]=1;
for(int i=1; i<32; i++)
a[i]=a[i-1]*2;
}
int main()
{
int n,m;
init();
while(~scanf("%d%d",&n,&m))
{
int maxx=0;
for(int i=0; i<n; i++)
scanf("%s%d",c[i].str,&c[i].x);
for(int i=0; a[i]<=m; i++)
{
int xx=a[i],yy=a[i]-1,zz=a[i]+1;
for(int j=0; j<n; j++)
{
//cout<<c[j].str<<" "<<c[j].x<<endl;
if(c[j].str[0]=='A')
{
xx=xx&c[j].x;
yy=yy&c[j].x;
zz=zz&c[j].x;
}
else if(c[j].str[0]=='O')
{
xx=xx|c[j].x;
yy=yy|c[j].x;
zz=zz|c[j].x;
}
else if(c[j].str[0]=='X')
{
xx=xx^c[j].x;
yy=yy^c[j].x;
zz=zz^c[j].x;
}
}
maxx=max(maxx,max(xx,max(yy,zz)));
}
printf("%d\n",maxx);
}
return 0;
}
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<cstdio>
#include<sstream>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include <ctype.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;
const int maxn=110000;
typedef long long ll;
const int INF=0x3f3f3f3f;
struct node
{
char c;
int num;
} a[100010];
int val_1[110],bin[110],val_0[110],max_val[110];
int solve(int n,int m,int max_num)
{
int i,j,sizes=0,ans=0;
for(i=0; (1<<i)<=max_num; i++)
{
val_1[i]=1<<i;
val_0[i]=0;
for(j=1; j<=m; j++)
{
if(a[j].c=='A') //AND
val_1[i]&=(a[j].num&(1<<i)),val_0[i]&=(a[j].num&(1<<i));
else if(a[j].c=='O') //OR
val_1[i]|=(a[j].num&(1<<i)),val_0[i]|=(a[j].num&(1<<i));
else if(a[j].c=='X') //XOR
val_1[i]^=(a[j].num&(1<<i)),val_0[i]^=(a[j].num&(1<<i));
}
max_val[i]=max(val_1[i],val_0[i]);
}
int max_sizes=0;
while(max_num)
max_sizes++,max_num>>=1;
while(n)
bin[++sizes]=n&1,n>>=1;
for(i=1; i<=max_sizes; i++)
{
if(bin[i]==0) ans+=val_0[i-1];
else
{
int t_ans=val_0[i-1];
for(j=1; j<i; j++) t_ans+=max_val[j-1];
ans=max(ans+val_1[i-1],t_ans);
}
}
return ans;
}
int main()
{
int n,m;
char s[10];
while(scanf("%d%d",&m,&n)!=EOF)
{
int max_num=0;
for(int i=1; i<=m; i++)
scanf("%s%d",s,&a[i].num),a[i].c=s[0],max_num=max(max_num,a[i].num);
printf("%d\n",solve(n,m,max_num));
}
return 0;
}
