http://acm.hdu.edu.cn/showproblem.php?pid=4768

貌似非常多人是用的二分

可是更好的做法貌似还是异或

对于第k个人。假设他接到偶数个传单。那么异或的结果还是0

就是说op记录全部收到传单的人次的总的异或值。那么由于仅仅有一个是收到奇数次。所以异或值就是他的编号，至于收到几次，在O(n)能够计算

//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;

#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const double pi = acos(-1.0);
const int INF = 100000000;
const int MAXN = 20010  +20;

int a[MAXN],b[MAXN],c[MAXN],n;

int main()
{
    while(~scanf("%d",&n)){
        int op=0,cnt=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d%d",&a[i],&b[i],&c[i]);
            for(int j=a[i];j<=b[i];j+=c[i])
                op^=j;
        }
        for(int i=0;i<n;i++){
            if(op>=a[i]  && op <=b[i] && (op-a[i])%c[i]==0)cnt++;
        }
        if(cnt%2==0)printf("DC Qiang is unhappy.\n");
        else printf("%d %d\n",op,cnt);
    }
    return 0;
}