Half thirds Study Notes

Front

No front

Bipartite

Half the maximum value of the subject is usually described using the minimum or maximum required minimum

Some specific idea is to keep a range is divided into two sections, divided into two sections continue to take legal

Constantly close to a very small range, the minimum range is the answer

But the point is usually either a legal or illegal, do not have points down

There are two binary model, type 1100 and type 0011 (a method, not legal 0)

A simple example to understand what

P1024 Cubic equation solver

This question is a typical 0011 type, to two points as the legitimate product is less than zero for some interval condition

If mid ~ r l = mid legitimate

Otherwise, r = mid

Then two time-card bit precision to

Code

#include <cstdio>
#include <algorithm>
#define maxn 500010

using namespace std;

const double eps = 1e-4;

double a, b, o, d, mid;

double c(double x)
{
    return x * x * x * a + x * x * b + x * o + d;
}

int main() {
    scanf("%lf%lf%lf%lf", &a, &b, &o, &d);
    double l, r;
    for(int i = -100; i <= 100; i++)
    {
        l = i;
        r = i + 1;
        if(c(l) == 0) 
        {
            printf("%.2f ", l);
            continue;
        }
        if(c(l) * c(r) < 0)
        {
            while(r - l > eps)//卡精度
            {
                mid = (l + r) / 2;
                if(c(mid) * c(r) <= 0) l = mid;//合法or不合法
                else r = mid;
            }
            printf("%.2f ", l);
        }
    }
    return 0;
}

A little difficult

P1182 series segments Section II

Half of the main difficulties in determining the legality

This question our mid maximum current

If the maximum number of groups in order to drop the sum <= m is legitimate

Otherwise illegal

Therefore, this type entitled 0011

#include <cstdio>
#include <algorithm>
#define maxn 500010

using namespace std;

int l, r, n, m, a[maxn], now, sum;

bool pd(int mid)
{
    now = mid;
    sum = 1;//初始化
    for(int i = 1; i <= n; i++)
    {
        if(now < a[i])//如果现在的组数数值最大值不够a[i],就开一个新的组
        {
            sum++;
            now = mid - a[i];
        }
        else now -= a[i];
    }
    return sum <= m;
}

int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        l = max(l ,a[i]);
        r += a[i];
    }
    int mid;
    while(l < r)
    {
        mid = (l + r) >> 1;
        if(pd(mid)) r = mid;
        else l = mid + 1;
    }
    printf("%d", l);
    return 0;
}

P2985 eating chocolate Chocolate Eating

This question refers to the mid value most unhappy happy day

Enumerate every day, today, when the value is less than happy eating a chocolate mid to determine when, if you still eat chocolate less than the mid to explain illegal

Note that the current record can not eat chocolate when looking for a minimum order value happy, because the program will all possible cases are down

We should be in the final again pitted pd, this time in order to write down

#include <cstdio>
#include <algorithm>
#include <cstring>
#define maxn 50010
typedef int int_ ;
#define int long long

using namespace std;

int val[maxn], n, d, l, r, vis[maxn], ans;

bool pd(int mid)
{
    int now = 0, last = 0;
    for(int j = 1; j <= d; j++)
    {
        last /= 2;
        while(last < mid)
        {
            last += val[++now];
            if(now > n) return 0;
            if(mid && mid == ans)vis[now] = j;
        }
    }
    return 1;
}

int_ main() {
    scanf("%lld%lld", &n, &d);
    for(int i = 1; i <= n; i++)
    {
        scanf("%lld", &val[i]);
        r += val[i];
    }
    int l = 0;
    while(l <= r)
    {
        int mid = (l + r) >> 1;
        if(pd(mid)) 
        {
            l = mid + 1;
            ans = mid;
        }
        else r = mid - 1;
    }
    pd(ans);//最后对答案跑一遍记录顺序
    printf("%lld\n", ans);
    for(int i = 1; i <= n; i++)
    {
        if(vis[i] == 0) vis[i] = d;
        printf("%lld\n", vis[i]);
    }
    return 0;
}

Thirds

#include <cstdio>
#include <algorithm>
#define maxn 500010

using namespace std;

const double eps = 1e-7;

int n;
double l, r, xs[110];

double f(double x)
{
    double ret = 0;
    for(int i = 0; i <= n; i++)
    {
        double cf = 1;
        for(int j = 1; j <= i; j++) cf *= x;
        ret += xs[i] *cf;
    }
    return ret;
}

int main() {
    scanf("%d%lf%lf", &n, &l, &r);
    for(int i = n; i >= 0; i--) scanf("%lf", &xs[i]);
    while(r - l > eps)
    {
        double mid = (l + r) / 2.0;
        double mmid = (mid + r) / 2.0;
        if(f(mid) > f(mmid)) r = mmid;
        else l = mid;
    }
    printf("%.5lf", l);
    return 0;
}

As the title. . . Thirds

We have continued to compare the size of the corresponding function and mid mmid see who greater

Application and unimodal function

（~~It can be used on the line~~）