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。

#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#include <cstdio>//sprintf islower isupper
#include <cstdlib>//malloc  exit strcat itoa system("cls")
#include <iostream>//pair
#include <fstream>//freopen("C:\Users\13606\Desktop\Input.txt","r",stdin);
#include <bitset>
//#include <map>
//#include<unordered_map>   http://acm.hdu.edu.cn/showproblem.php?pid=4149
#include <vector>
#include <stack>
#include <set>
#include <string.h>//strstr substr strcat
#include <string>
#include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
#include <cmath>
#include <deque>
#include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
#include <vector>//emplace_back
//#include <math.h>
#include <cassert>
#include <iomanip>
//#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
#include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
//******************
clock_t __START,__END;
double __TOTALTIME;
void _MS(){__START=clock();}
void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
//***********************
#define rint register int
#define fo(a,b,c) for(rint a=b;a<=c;++a)
#define fr(a,b,c) for(rint a=b;a>=c;--a)
#define mem(a,b) memset(a,b,sizeof(a))
#define pr printf
#define sc scanf
#define ls rt<<1
#define rs rt<<1|1
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef unsigned long long ull;
typedef long long ll;
typedef double db;
const db E=2.718281828;
const db PI=acos(-1.0);
const ll INF=(1LL<<60);
const int inf=(1<<30);
const db ESP=1e-9;
const int mod=(int)1e9+7;
const int N=(int)1e6+10;

int x[10];
int t[10][35];
void in(int x,int to)
{
    for(int i=0;i<=30;++i)
        t[to][i]=(x>>i)&1;
}

void solve()
{
    for(int i=1;i<=8;++i)sc("%d",&x[i]),in(x[i],i);
    int sum;
    sc("%d",&sum);
    in(sum,9);
    int now=0,st=1;
    bool is[35];mem(is,0);
    for(int i=0;i<=30;++i)
    {
        for(int j=1;j<=8;++j)
                now+=t[j][i]*st;
        if(((now>>i)&1)!=((sum>>i)&1))
        {
            is[i]=1;
            for(int j=1;j<=8;++j)
            {
                now-=t[j][i]*st;
                t[j][i]^=1;
                now+=t[j][i]*st;
            }
        }
        st*=2;
    }
    int ans=0;
    for(int i=0;i<=30;++i)if(is[i])ans|=1<<i;
    pr("%d
",ans);
}

int main()
{
    int T;
    sc("%d",&T);
    while(T--)solve();
    return 0;
}

/**************************************************************************************/