HDU 4768 Flyer 二分

因为保证只有一个数为奇数,所以所有区间内满足的数的总数一定是奇数

且包含这个数的所有区间内 能满足条件的数的个数和也一定是奇数

那么二分这个数...在所有的区间内查找满足的条件...代码不到20行

/********************* Template ************************/
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define EPS         1e-8
#define DINF        1e15
#define MAXN        20100
#define LINF        1LL << 60
#define MOD         95041567//1000000007
#define INF         0x7fffffff
#define PI          3.14159265358979323846
#define lson            l,m,rt<<1
#define rson            m+1,r,rt<<1|1
#define BUG             cout<<" BUG! "<<endl;
#define LINE            cout<<" ------------------ "<<endl;
#define FIN             freopen("in.txt","r",stdin);
#define FOUT            freopen("out.txt","w",stdout);
#define mem(a,b)        memset(a,b,sizeof(a))
#define FOR(i,a,b)      for(int i = a ; i < b ; i++)
#define read(a)         scanf("%d",&a)
#define read2(a,b)      scanf("%d%d",&a,&b)
#define read3(a,b,c)    scanf("%d%d%d",&a,&b,&c)
#define write(a)        printf("%d\n",a)
#define write2(a,b)     printf("%d %d\n",a,b)
#define write3(a,b,c)   printf("%d %d %d\n",a,b,c)
#pragma comment         (linker,"/STACK:102400000,102400000")
template<class T> inline T L(T a)       {return (a << 1);}
template<class T> inline T R(T a)       {return (a << 1 | 1);}
template<class T> inline T lowbit(T a)  {return (a & -a);}
template<class T> inline T Mid(T a,T b) {return ((a + b) >> 1);}
template<class T> inline T gcd(T a,T b) {return b ? gcd(b,a%b) : a;}
template<class T> inline T lcm(T a,T b) {return a / gcd(a,b) * b;}
template<class T> inline T Min(T a,T b) {return a < b ? a : b;}
template<class T> inline T Max(T a,T b) {return a > b ? a : b;}
template<class T> inline T Min(T a,T b,T c)     {return min(min(a,b),c);}
template<class T> inline T Max(T a,T b,T c)     {return max(max(a,b),c);}
template<class T> inline T Min(T a,T b,T c,T d) {return min(min(a,b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d) {return max(max(a,b),max(c,d));}
template<class T> inline T exGCD(T a, T b, T &x, T &y){
    if(!b) return x = 1,y = 0,a;
    T res = exGCD(b,a%b,x,y),tmp = x;
    x = y,y = tmp - (a / b) * y;
    return res;
}
//typedef long long LL;    typedef unsigned long long ULL;
typedef __int64 LL;      typedef unsigned __int64 ULL;
/*********************   By  F   *********************/
LL a[MAXN],b[MAXN],c[MAXN];
int n;
bool solve(LL m){
    LL ret = 0;
    for(int i = 0 ; i < n ; i++){
        if(m >= b[i]) ret += (b[i] - a[i]) / c[i] + 1;
        else if(m >= a[i]) ret += (m - a[i]) / c[i] + 1;
    }
    if(ret&1) return true;
    return false;
}
int main(){
    while(~scanf("%d",&n)){
        for(int i = 0 ; i < n ; i++) scanf("%I64d%I64d%I64d",&a[i],&b[i],&c[i]);
        LL l = 0 , r = 1LL<<32;
        while(l < r){
            LL m = (l+r)/2;
            if(solve(m)) r = m;
            else l = m + 1;
        }
        LL cnt = 0 ;
        for(int i = 0 ; i < n ; i++)
            if(l <= b[i] && l >= a[i] && (l-a[i])%c[i] == 0) cnt++;
        if(cnt) printf("%I64d %I64d\n",l,cnt);
        else printf("DC Qiang is unhappy.\n");
    }
    return 0;
}