hdu 4768 Flyer (二分)

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

 

算法思路：最核心的就是找到前n个人的传单数总和可以二分。

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 20002;

long long A[maxn],B[maxn],C[maxn];
int N;

int main()
{
    //freopen("E:\\acm\\input.txt","r",stdin);
    while(cin>>N)
    {
        long long l = 1e12;
        long long r = 0;
        for(int i=1; i<=N; i++)
        {
            scanf("%I64d %I64d %I64d",&A[i],&B[i],&C[i]);
            l = min(l,A[i]);
            r = max(r,B[i]);
        }
        long long sum = 0;
        for(int i=1; i<=N; i++)
        {
            sum += (B[i]-A[i])/C[i] + 1;
        }

        if(sum%2 == 0)
        {
            printf("DC Qiang is unhappy.\n");
        }
        else
        {

            while(l < r)
            {
                long long mid = (l+r)/2;
                sum = 0;
                for(int i=1; i<=N; i++)
                {
                    int temp = min(B[i],mid)-A[i];
                    if(temp >= 0)
                        sum += temp/C[i]+1;
                }
                if(sum%2) r = mid;
                else      l = mid+1;
            }

            int num = 0;
            for(int i=1; i<=N; i++)
            {
                if(l>=A[i] && (l-A[i])%C[i] == 0 && l <= B[i])
                    num++;
            }
            printf("%lld %d\n",l,num);
        }
    }
}