hdu 4768 Flyer 二分

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

思路：

解题关键是奇数+偶数=奇数，然后我们就是枚举奇数位置（奇数为就一个或者0个），然后计算左边的和是否为奇数，如果是奇数，那么该点就存在与左边，否则存在于右边

 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <time.h>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define keyTree (chd[chd[root][1]][0])
#define Read()  freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);
#define random(x) (rand()%x)

#define M 100007
#define N 20007

using namespace std;

int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1};

const int inf = 0x7f7f7f7f;
const int mod = 1000000007;

const double eps = 1e-8;
const int R = 100007;

int A[N],B[N],C[N];
int n;

ll solve(int mid)
{
    ll sum = 0;
    int limt;
    for (int i = 0; i < n; ++i)
    {
        if(mid < A[i])  continue;
        limt = min(mid, B[i]);
        if (C[i] == 0) sum += 1;
        else
        {
            sum += (limt - A[i])/C[i] + 1;
        }
    }
    return sum;
}
int main()
{
    while (~scanf("%d",&n))
    {
        int top = 0;
        for (int i = 0; i < n; ++i)
        {
            cin>>A[i]>>B[i]>>C[i];
            top = max(top,B[i]);
        }
        int l = 1, r = top;
        int ans = 0;
        while (l <= r)
        {
            int mid = ((ll)l + (ll)r)>>1;
            ll sum = solve(mid);
            if ((sum & 1) == 1)
            {
                ans = mid;
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        if (ans == 0) printf("DC Qiang is unhappy.\n");
        else
        {
            int cnt = 0;
            for (int i = 0; i < n; ++i)
            {
                if (ans <= B[i])
                {
                    if (C[i] == 0){
                         if (ans == A[i]) cnt++;
                    } else {
                        if (ans >= A[i] && (ans - A[i])%C[i] == 0) cnt++;
                    }
                }
            }
            printf("%d %d\n",ans,cnt);
        }
    }
    return 0;
}