加法和XOR的搞搞--二进制（HDU4149 Magic Potion）

Posted 2021-03-09 --hpy-7m

题意：http://acm.hdu.edu.cn/showproblem.php?pid=4149

已知九个数：x1 xor m，x2 xor m......x8 xor m，(x1+x2+....+x8) xor m；求出m

思路：https://blog.csdn.net/AC_0_summer/article/details/46975751?utm_source=distribute.pc_relevant.none-task

差不多了。

 1 #define ios ios_base::sync_with_stdio(0); cin.tie(0);
 2 #include <cstdio>//sprintf islower isupper
 3 #include <cstdlib>//malloc  exit strcat itoa system("cls")
 4 #include <iostream>//pair
 5 #include <fstream>//freopen("C:\Users\13606\Desktop\Input.txt","r",stdin);
 6 #include <bitset>
 7 //#include <map>
 8 //#include<unordered_map>   http://acm.hdu.edu.cn/showproblem.php?pid=4149
 9 #include <vector>
10 #include <stack>
11 #include <set>
12 #include <string.h>//strstr substr strcat
13 #include <string>
14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
15 #include <cmath>
16 #include <deque>
17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
18 #include <vector>//emplace_back
19 //#include <math.h>
20 #include <cassert>
21 #include <iomanip>
22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
25 //******************
26 clock_t __START,__END;
27 double __TOTALTIME;
28 void _MS(){__START=clock();}
29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
30 //***********************
31 #define rint register int
32 #define fo(a,b,c) for(rint a=b;a<=c;++a)
33 #define fr(a,b,c) for(rint a=b;a>=c;--a)
34 #define mem(a,b) memset(a,b,sizeof(a))
35 #define pr printf
36 #define sc scanf
37 #define ls rt<<1
38 #define rs rt<<1|1
39 typedef pair<int,int> PII;
40 typedef vector<int> VI;
41 typedef unsigned long long ull;
42 typedef long long ll;
43 typedef double db;
44 const db E=2.718281828;
45 const db PI=acos(-1.0);
46 const ll INF=(1LL<<60);
47 const int inf=(1<<30);
48 const db ESP=1e-9;
49 const int mod=(int)1e9+7;
50 const int N=(int)1e6+10;
51 
52 int x[10];
53 int t[10][35];
54 void in(int x,int to)
55 {
56     for(int i=0;i<=30;++i)
57         t[to][i]=(x>>i)&1;
58 }
59 
60 void solve()
61 {
62     for(int i=1;i<=8;++i)sc("%d",&x[i]),in(x[i],i);
63     int sum;
64     sc("%d",&sum);
65     in(sum,9);
66     int now=0,st=1;
67     bool is[35];mem(is,0);
68     for(int i=0;i<=30;++i)
69     {
70         for(int j=1;j<=8;++j)
71                 now+=t[j][i]*st;
72         if(((now>>i)&1)!=((sum>>i)&1))
73         {
74             is[i]=1;
75             for(int j=1;j<=8;++j)
76             {
77                 now-=t[j][i]*st;
78                 t[j][i]^=1;
79                 now+=t[j][i]*st;
80             }
81         }
82         st*=2;
83     }
84     int ans=0;
85     for(int i=0;i<=30;++i)if(is[i])ans|=1<<i;
86     pr("%d
",ans);
87 }
88 
89 int main()
90 {
91     int T;
92     sc("%d",&T);
93     while(T--)solve();
94     return 0;
95 }
96 
97 /**************************************************************************************/