HDU4768：Flyer [ 二分的奇妙应用 好题 ]

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Flyer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1718    Accepted Submission(s): 622

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

 

 

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题意及题解转自：http://blog.csdn.net/libin56842/article/details/25741317

题意：n个社团派发传单，有a,b,c三个参数，派发的规则是，派发给序号为a，a+c....a+k*c，序号要求是小于等于b

这其中，有一个学生只收到了奇数传单，要求找出这个学生的编号与得到的传单数目

 

思路：如果使用异或运算，也还是比较简单的，但是这样的话所花费的时间就比较长，正确的做法是使用二分

使用二分来划分区间，由于是每个社团得到的序列都是等差数列，所以我们很容易能得到区间派发的传单数

如果是奇数，那么所求的人肯定在左区间，否则在右区间，这样二分下去找到答案

 

 

12940329

2015-02-12 17:35:01

Accepted

4768

78MS

1756K

1933 B

G++

czy

 

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<string>

#define N 20005
#define M 205
#define mod 10000007
//#define p 10000007
#define mod2 1000000000
#define ll long long
#define LL long long
#define eps 1e-6
#define inf 100000000
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b)

using namespace std;

int n;
ll a[N],b[N],c[N];
ll num[N];
ll ans;
ll ma;
ll cc;

void ini()
{
    int i;
    ma=0;
    for(i=1;i<=n;i++){
        scanf("%I64d%I64d%I64d",&a[i],&b[i],&c[i]);
        ma=max(ma,b[i]);
        num[i]=(b[i]-a[i])/c[i]+1;
    }
}

ll ok(ll l,ll r)
{
    int i;
    ll re;
    re=0;
    ll t1,t2;
    l--;
    for(i=1;i<=n;i++){
        if(l>b[i] || r<a[i]) continue;
        if(l<a[i]){
            t1=0;
        }
        else{
            t1=(l-a[i])/c[i]+1;
        }
        if(r>b[i]){
            t2=num[i];
        }
        else{
            t2=(r-a[i])/c[i]+1;
        }
        re+=t2-t1;
    }
    if(re%2==1) return re;
    else return 0;
}

void solve()
{
    ll l,r,mid;
    l=0;r=ma;
    while(l<r)
    {
        mid=(l+r)/2;
        cc=ok(l,mid);
        if(cc!=0)
        {
            r=mid;
        }
        else{
            l=mid+1;
        }
    }
    ans=l;
    cc=ok(l,l);
}

void out()
{
    if(cc==0){
        printf("DC Qiang is unhappy.\n");
    }
    else
        printf("%I64d %I64d\n",ans,cc);
}

int main()
{
    //freopen("data.in","r",stdin);
    //freopen("data.out","w",stdout);
    //scanf("%d",&T);
    //for(int ccnt=1;ccnt<=T;ccnt++)
    //while(T--)
    //scanf("%d%d",&n,&m);
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0) break;
        ini();
        solve();
        out();
    }
    return 0;
}