HDU 4768: Flyer

题意：

有N个社团，每个社团三个属性A,B,C，表示会向编号A+k*C的同学发传单(k=0,1,2...  && A+k*C <= B)。题目保证最多有一个人收到的传单数是奇数。求如果有人收到传单是奇数，输出编号和他收到的传单数。

思路：

观察最后情况，可以发现，要么每个人都是偶数。要么有一个是奇数。观察其前缀和，发现奇数那个人之前的前缀和都是偶数，之后都是奇数。所以，二分之。

代码：

（上交大牛代码……）

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cmath>
#include <iostream>
#include <algorithm>

using namespace std;

int n, a[100001][3];
long long sum;

bool check(int now){
    long long sum = 0;
    for (int i = 1; i <= n; i++)
        if (now >= a[i][0])
        {
            int Right = min(a[i][1], now);
            if (a[i][2]) sum += (long long)(Right - a[i][0]) / a[i][2] + 1;
            else sum++;
        }
    return(sum & 1);
}

int read(){
    char ch;
    for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar());
    int cnt = 0;
    for (; ch >= '0' && ch <= '9'; ch = getchar()) cnt = cnt * 10 + ch - '0';
    return(cnt);
}


int main(){
    //freopen("j.in", "r", stdin);
    //freopen("j.out", "w", stdout);
    for (;;)
    {
        if (scanf("%d", &n) != 1) return 0;
        sum = 0;
        int Max = 0;
        for (int i = 1; i <= n; i++) 
        {
            a[i][0] = read(); a[i][1] = read(); a[i][2] = read();
            if (a[i][2]) sum += (long long)(a[i][1] - a[i][0]) / a[i][2] + 1;
            else sum++;
            Max = max(Max, a[i][1]);
        }
        if (!(sum & 1))
        {
            printf("DC Qiang is unhappy.\n");
            continue;
        }
        long long Left = 0, Right = Max, Mid = (Left + Right) >> 1;
        while (Left + 1 < Right)
        {
            if (check(Mid)) Right = Mid;
            else Left = Mid;
            Mid = (Left + Right) >> 1;
        }
        printf("%d", Right);
        int cnt = 0;
        for (int i = 1; i <= n; i++)
            if (Right >= a[i][0] && Right <= a[i][1] && !((Right - a[i][0]) % a[i][2])) ++cnt;
        printf(" %d\n", cnt);
    }
}