Flyer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 386    Accepted Submission(s): 127

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  

Source
2013 ACM/ICPC Asia Regional Changchun Online  

Recommend liuyiding  

因为最多一个奇数的，所以总数是奇数的。

二分区间就可以了。

比赛时候二分的 l,r都是int存的，导致算mid = (l+r)/2的时候溢出了，TAT

这样TLE了好久，我艹。。。。。

水题一发

 1 /* ***********************************************
 2 Author        :kuangbin
 3 Created Time  :2013/9/28 星期六 13:00:24
 4 File Name     :2013长春网络赛\1010.cpp
 5 ************************************************ */
 6 
 7 #pragma comment(linker, "/STACK:1024000000,1024000000")
 8 #include <stdio.h>
 9 #include <string.h>
10 #include <iostream>
11 #include <algorithm>
12 #include <vector>
13 #include <queue>
14 #include <set>
15 #include <map>
16 #include <string>
17 #include <math.h>
18 #include <stdlib.h>
19 #include <time.h>
20 using namespace std;
21 
22 const int MAXN = 20010;
23 long long A[MAXN];
24 long long B[MAXN];
25 long long C[MAXN];
26 int n;
27 inline long long calc(long long  start,long long add,long long end)
28 {
29     return (end-start+1 + add-1)/add;
30 }
31 inline bool check(long long l,long long r)
32 {
33     long long sum = 0;
34     for(int i = 0;i < n;i++)
35     {
36         long long end = min(B[i],r);
37         if(end < A[i])continue;
38         if(l >end)continue;
39         long long s1;
40         s1 = calc(A[i],C[i],end);
41         if(l <= A[i])sum += s1;
42         else sum += s1 - calc(A[i],C[i],l-1);
43 
44     }
45     if(sum % 2 == 0)return false;
46     return true;
47 }
48 long long ttt(long long l,long long r)
49 {
50     long long sum = 0;
51     for(int i = 0;i < n;i++)
52     {
53         long long end = min(B[i],r);
54         if(end < A[i])continue;
55         if(l > end)continue;
56         long long s1,s2;
57         s1 = calc(A[i],C[i],end);
58         if(l <= A[i])sum += s1;
59         else sum += s1 - calc(A[i],C[i],l-1);
60 
61     }
62     return sum;
63 }
64 int main()
65 {
66     //freopen("in.txt","r",stdin);
67     //freopen("out.txt","w",stdout);
68     while(scanf("%d",&n) == 1 && n)
69     {
70         long long l = 100000000000LL, r = 0;
71         for(int i = 0;i < n;i++)
72         {
73             scanf("%I64d%I64d%I64d",&A[i],&B[i],&C[i]);
74             l = min(l,A[i]);
75             r = max(r,B[i]);
76         }
77         if(check(l,r) == false)
78         {
79             printf("DC Qiang is unhappy.\n");
80             continue;
81         }
82         while(l < r)
83         {
84             long long mid = (l+r)/2;
85             if(check(l,mid))
86             {
87                 r = mid;
88             }
89             else l = mid+1;
90         }
91         printf("%d %d\n",(int)l,(int)ttt(l,l));
92     }
93     return 0;
94 }