A. Short Program
很巧妙的处理方式,用0和1023来记录
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 int main(){ 5 int n; 6 scanf("%d", &n); 7 char s[3]; 8 int op; 9 int a = 0, b = 1023; 10 for(int i = 0; i < n; i++){ 11 scanf("%s %d", s, &op); 12 if(s[0] == '|') a |= op, b |= op; 13 if(s[0] == '&') a &= op, b &= op; 14 if(s[0] == '^') a ^= op, b ^= op; 15 } 16 int u = 0, v = 0; 17 for(int i = 0; i < 10; i++){ 18 int x = 1<<i; 19 if(a&x && b&x) u |= x; 20 if(a&x && ~b&x) v |= x; 21 if(~a&x && ~b&x) u |= x, v |= x; 22 } 23 printf("2 | %d ^ %d ",u, v); 24 }