2020-09-15 [TJOI2007]路标设置

https://www.luogu.com.cn/problem/P3853

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <sstream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;

//template<typename DataType>
//DataType qmi(DataType m, int k)
//{
//    DataType res = 1, t = m;
//    while (k)
//    {
//        if (k&1) res = res * t;
//        t = t * t;
//        k >>= 1;
//    }
//    return res;
//}


int qmi(int m, int k)
{
    int res = 1, t = m;
    while (k)
    {
        if (k&1) res = res * t;
        t = t * t;
        k >>= 1;
    }
    return res;
}

int read(){
    int x = 0,f = 1;
    char c = getchar();
    while (c<'0'||c>'9') {
        if (c=='-') {
            f = -1;
        }
        c = getchar();
    }
    while (c>='0'&&c<='9') {
        x = x*10+c-'0';
        c = getchar();
    }
    return x*f;
}

#define fi(a,b) for(int i = a; i <= b; i++)
#define fj(a,b) for(int j = a; j >= b; j--)

struct priceAndCnt{
    int price,cnt;
};

void quickSort(priceAndCnt *a,int left,int right){
    int i,j;
    
    priceAndCnt temp,t;
    
    temp = a[(left+right)/2];//基准值
    
    i = left;
    j = right;
    
    while(i<=j){
        while (a[j].price > temp.price) {
            j--;
        }
        
        while (a[i].price < temp.price) {
            i++;
        }
        
        if (i<=j) {
            t = a[i];
            a[i] = a[j];
            a[j] = t;
            //继续下一步
            i++;
            j--;
        }
    }
    
    if(left<j)quickSort(a,left, j);//继续分治
    if(i<right)quickSort(a,i, right);//继续分治
}


int L,N,K;
int arr[100005];

int Judge(int mid)
{
    int m = 0;
    fi(2,N){
        if(arr[i]-arr[i-1]>=mid)
        {
            m+=(arr[i]-arr[i-1])/mid;
            if((arr[i]-arr[i-1])%mid==0)
                m--;
        }
    }
    if(m > K)
        return 0;
    return 1;
}

int main()
{
    L = read();
    N = read();
    K = read();
    
    fi(1, N){
        arr[i] = read();
    }
    
    int l = 0;//l和r分别代表二分的左边界和右边界
    int r = L;
    while (l < r){//非递归式二分正常向写法，可理解为一般框架
        int mid = (l + r) / 2;//这再看不出是啥意思可以退群了
        if (Judge(mid) == 0){//带入judge函数判断当前解是不是可行解
            l = mid + 1;
        } else {
            r = mid;
        }
    }
    cout << l << endl;//最后的ans绝对是最优解
    return 0;
}
/*
101 2 1
0 101
============
51
*/