BZOJ 十连测 二进制的世界

SOL：

   我们可以分块啊·。

#include<bits/stdc++.h>
using namespace std;
#define sight(c) ('0'<=c&&c<='9')
char ch[10]; int f[277][277],g[277][277],v[277],k,ans,cnt,n,op,x;
inline void read(int &x){
    static char c;
    for (c=getchar();!sight(c);c=getchar());
    for (x=0;sight(c);c=getchar())x=x*10+c-48;
}
void write(int x){if (x<10) {putchar('0'+x); return;} write(x/10); putchar('0'+x%10);}
inline void writeln(int x){ if (x<0) putchar('-'),x*=-1; write(x); putchar('
'); }
inline void writel(int x){ if (x<0) putchar('-'),x*=-1; write(x); putchar(' '); }
inline int fun(int x,int y) {
     if (ch[0]=='a') return x&y;
     if (ch[0]=='o') return x|y;
     return x^y;
}
void add(int x){
    int A=x>>8,B=x&255; v[A]=1;
    for (int i=0;i<256;i++) {
        k=fun(i,B);
        if (k>f[A][i]) f[A][i]=k,g[A][i]=1;
        else if (k==f[A][i]) g[A][i]++;
    }
}
void ask(int x) {
    int A=x>>8,B=x&255;
    ans=cnt=0;
    for (int i=0;i<256;i++) if (v[i]){
        k=fun(i,A)<<8|f[i][B];
        if (k>ans) ans=k,cnt=g[i][B];
        else if (k==ans) cnt+=g[i][B];
    }
}
signed main () {
    freopen("binary.in","r",stdin);
    freopen("binary.out","w",stdout);
    scanf("%d%s%d",&n,ch,&op);
    for (int i=1;i<=n;i++) {
       read(x);
       if (i!=1) { ask(x);
         if (op) writel(ans),writeln(cnt);
         else writeln(ans);
       }
       add(x);
    }
}