Bzoj 3668 noi2014 getting up difficult syndrome greedy

A bunch of nonsense ..

Given N operations (and, Or, different or ~ Select a number in m to maximize the value after N operations.

A great deal of water, such as soil, such as iron ..

This Nima is simply a question of water. We can discuss the following three situations when selecting each digit of a number:

1. When this bit is set to 0, the result obtained after N operations is set to 1. This is the ideal case. You must select 0.

2. Case 1 does not meet the requirement. If this parameter is set to 1, 1 is obtained after N operations and the value after 1 does not exceed M. In this case, we also select 1.

3. If the preceding two conditions are not met, the bitwise 0 must be smaller than the bitwise 1, which is more difficult than M.

So the question goes.

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define M 100100using namespace std;struct abcd{    int p,x;    int cross(int y)    {        if(p==0)            return x&y;        if(p==1)            return x|y;        return x^y;    }}a[M];int n,m;char s[100];int cross(int x){    int i;    for(i=1;i<=n;i++)        x=a[i].cross(x);    return x;}int main(){              int i,ans=0,now;    cin>>n>>m;    for(i=1;i<=n;i++)    {        scanf("%s",s);        if(s[0]=='A')            a[i].p=0;        else if(s[0]=='O')            a[i].p=1;        else            a[i].p=2;        scanf("%d",&a[i].x);    }    for(now=1;now<=m;now<<=1);    for(now>>=1;now;now>>=1)    {        if(cross(0)&now)            continue;        if(ans+now<=m&&cross(now)&now)            ans+=now;    }    printf("%d\n",cross(ans));}

Bzoj 3668 noi2014 getting up difficult syndrome greedy