加法和XOR的搞搞--二进制（HDU4149 Magic Potion）
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\n",ans); 87 } 88 89 int main() 90 { 91 int T; 92 sc("%d",&T); 93 while(T--)solve(); 94 return 0; 95 } 96 97 /**************************************************************************************/
