HDU 4768 Flyer(二分)

Flyer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1445    Accepted Submission(s): 510

Problem Description

The new semester begins! Different kinds of student societies are all trying to advertise themselves, by giving flyers to the students for introducing the society. However, due to the fund shortage, the flyers of a society can only be distributed to a part of the students. There are too many, too many students in our university, labeled from 1 to 2^32. And there are totally N student societies, where the i-th society will deliver flyers to the students with label A_i, A_i+C_i,A_i+2*C_i,…A_i+k*C_i (A_i+k*C_i<=B_i, A_i+(k+1)*C_i>B_i). We call a student "unlucky" if he/she gets odd pieces of flyers. Unfortunately, not everyone is lucky. Yet, no worries; there is at most one student who is unlucky. Could you help us find out who the unfortunate dude (if any) is? So that we can comfort him by treating him to a big meal!

 

Input

There are multiple test cases. For each test case, the first line contains a number N (0 < N <= 20000) indicating the number of societies. Then for each of the following N lines, there are three non-negative integers A_i, B_i, C_i (smaller than 2^31, A_i <= B_i) as stated above. Your program should proceed to the end of the file.

 

Output

For each test case, if there is no unlucky student, print "DC Qiang is unhappy." (excluding the quotation mark), in a single line. Otherwise print two integers, i.e., the label of the unlucky student and the number of flyers he/she gets, in a single line.

 

Sample Input

2 1 10 1 2 10 1 4 5 20 7 6 14 3 5 9 1 7 21 12

 

Sample Output

1 1 8 1

题意：给定n个分传单，每次从a分到b间隔为c，保证最多一个人分传单为奇数，问这个人是第几个和传单数，如果不存在输出DC Qiang is unhappy.

思路：二分，计算总和，总和为偶数不可行，然后去二分奇数人的位置，计算左边一块是否是奇数，如果是就说明该人在左边往左缩，否则就往右。

代码:

#include 
#include 
#include 
const long long INF = (1LL<<31);
#include 
using namespace std;

const int N = 20005;
int n;
long long sum, a[N], b[N], c[N], l, r;

long long judge(long long mid) {
    long long sum = 0;
    for (int i = 0; i < n; i++) {
	long long br = min(mid , b[i]);
	if (br >= a[i])
	    sum += (br - a[i]) / c[i] + 1;
    }
    return sum % 2;
}

bool solve() {
    if (sum % 2 == 0) return false;
    while (l < r) {
	long long mid = (l + r) / 2;
	if (judge(mid))
	    r = mid;
	else
	    l = mid + 1;
    }
    int num = 0;
    for (int i = 0; i < n; i++) {
	if (l >= a[i] && l <= b[i]) {
	    if ((l - a[i]) % c[i] == 0)
		num++;
	}
    }
    cout << l << ' ' << num << endl;
    return true;
}

int main() {
    while (scanf("%d", &n) == 1) {
	sum = 0; l = INF; r = 0;
	for (int i = 0; i < n; i++) {
	    cin >> a[i] >> b[i] >> c[i];
	    sum += (b[i] - a[i]) / c[i] + 1;
	    l = min(l, a[i]);
	    r = max(r, b[i]);
	}
	if (!solve())
	    cout << "DC Qiang is unhappy." << endl;
    }
    return 0;
}