2015
09-17

# Flyer

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!

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.

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.

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

1 1
8 1

n个最多2W个社团发传单，因为传单不够，所有只能按照学生学号间隔着发……  比如发给A_i,A-i+k,A_i+2*k  ,只要不大于B_I就好；

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

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn=20010;
long long A[maxn],B[maxn],C[maxn];
long long cal(long long mid,int n){
long long sum=0;
for (int i=0;i<n;i++){
long long up=min(mid,B[i]);
if (up>=A[i])
sum+=(up-A[i])/C[i]+1;
}
//    cout<<mid<<" "<<sum<<endl;
return sum;
}
int main()
{
int n;
while (scanf("%d",&n)==1)
{
for (int i=0;i<n;i++){
cin>>A[i]>>B[i]>>C[i];
}
long long L=0,R=1LL<<31;
while (L<R){
long long mid=(L+R)/2;
if (cal(mid,n)%2) R=mid;
else L=mid+1;
}
if (L==1LL<<31)
cout<<"DC Qiang is unhappy."<<endl;
else {
while (cal(L,n)%2==0) L--;
cout<<L<<' '<<cal(L,n)-cal(L-1,n)<<endl;
}
}
return 0;
}

1. 宾得（潘泰克斯）机械单反，毕竟完全的机械机械结构嘛，不像现在数码单反一个快门组件可以用集成电路板控制，得由无数的零件来控制。举个不恰当的例子，就像内燃机机与电机。嗯或许电子表和机械手表来对比更为贴切。