HDU2199,HDU2899,HDU1969,HDU2141--（简单二分）

二分是一种很有效的减少时间开销的策略, 我觉得单列出二分专题有些不太合理, 二分应该作为一中优化方法来考虑

这几道题都是简单的使用了二分方法优化, 二分虽然看似很简单, 但一不注意就会犯错. 在写二分时, 会遇到很多选择题, 很多"分叉路口", 要根据实际情况选择合适的"路"

HDU2199

Can you solve this equation?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1195 Accepted Submission(s): 565

Problem Description

Now,given the equation 8*x^4 + 7*x^3 + 2*x^2 + 3*x + 6 == Y,can you find its solution between 0 and 100;
Now please try your lucky.

Input

The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has a real number Y (fabs(Y) <= 1e10);

Output

            For each test case, you should just output one real number(accurate up to 4 decimal places),which is the solution of the equation,or “No solution!”,if there is no solution for the equation between 0 and 100.

Sample Input

2
100
-4

Sample Output

1.6152
No solution!

Author

Redow

Recommend

 

注意看AC代码,循环的条件是 l+eps<r ,不能用 f(mid)==res 否则会一直循环下去(注意看循环内部, 有可能边界(l或r)就为正确的值, 这样的话永远不会搜索到)

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<iomanip>
#define INF 0x7ffffff
#define MAXN 10000
using namespace std;
const double eps=1e-10;
double js(double x)
{
    return 8.0*x*x*x*x + 7.0*x*x*x + 2.0*x*x + 3.0*x + 6.0;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
    #endif
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    double y;
    double x;
    double r,l;
    int t;
    cin>>t;
    while(t--){
        cin>>y;
        if(y<6||y>js(100)){
            cout<<"No solution!"<<endl;
        }
        else{
            l=0; r=100;
            double mid,res;
            while(l+eps<r){
                mid=(l+r)/2;
                if(y<js(mid)){
                    r=mid;
                }
                else{
                    l=mid;
                }
            }
            cout<<fixed<<setprecision(4)<<mid<<endl;
        }
    }
    return 0;
}

HDU2899

Strange fuction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 666 Accepted Submission(s): 549

Problem Description

Now, here is a fuction:
  F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
Can you find the minimum value when x is between 0 and 100.

 

Input

The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has only one real numbers Y.(0 < Y <1e10)

 

Output

            Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.

 

Sample Input

2
100
200

 

Sample Output

-74.4291
-178.8534

 

Author

Redow

 

Recommend

lcy

和上一道题一模一样,只不过这道题目拐了个弯, 需要进行求导

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<iomanip>
#define INF 0x7ffffff
#define MAXN 10000
using namespace std;
const double eps=1e-10;
double fd(double x)
{
    return 42*pow(x,6)+48*pow(x,5)+21*pow(x,2)+10*x;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
    #endif
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int t;
    double y;
    double l,r,mid;
    cin>>t;
    while(t--){
        cin>>y;
        l=0;
        r=100;
        while(r-l>eps){
            mid=(l+r)/2;
            if(fd(mid)>y){
                r=mid;
            }
            else l=mid;
        }
        cout<<fixed<<setprecision(4)<<6*pow(mid,7)+8*pow(mid,6)+7*pow(mid,3)+5*mid*mid-y*mid<<endl;
    }
}

HDU1969

Pie

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 939 Accepted Submission(s): 348

Problem Description

My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.

My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.

What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.

 

Input

One line with a positive integer: the number of test cases. Then for each test case:
---One line with two integers N and F with 1 <= N, F <= 10 000: the number of pies and the number of friends.
---One line with N integers ri with 1 <= ri <= 10 000: the radii of the pies.

 

Output

For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10^(-3).

 

Sample Input

3
3 3
4 3 3
1 24
5
10 5
1 4 2 3 4 5 6 5 4 2

 

Sample Output

25.1327
3.1416
50.2655

 

Source

NWERC2006

 

Recommend

wangye

这道题没有固定的函数关系, 每一个待定的面积值都要去除每块pie的面积, 为了加快搜索到结果的速度, 可以采用二分的方法(可分成的块数大体上是相对于面积递增的)

代码

    #include <cstdio>  
    #include <cstring>  
    #include <algorithm>  
    #include <cmath>  
    using namespace std ;  
    #define PI 3.1415926535898  
    #define eqs 1e-5  
    double s[11000] ;  
    int n , m ;  
    double f(double x)  
    {  
        int k = (x+eqs) * 10000 ;  
        x = k * 1.0 / 10000 ;  
        return x ;  
    }  
    int solve(double x)  
    {  
        int i , j , num = 0 ;  
        for(i = n-1 ; i >= 0 && (s[i]-x) > eqs ; i--)  
        {  
            j = s[i] / x ;  
            num += j ;  
            if( num >= m+1 )return 1 ;  
        }  
        if( num >= m+1 ) return 1 ;  
        return 0 ;  
    }  
    int main()  
    {  
        int t , i , k ;  
        double low , mid , high , last ;  
        while( scanf("%d", &t) != EOF )  
        {  
            while(t--)  
            {  
                scanf("%d %d", &n, &m) ;  
                for(i = 0 ; i < n ; i++)  
                    scanf("%lf", &s[i]) ;  
                sort(s,s+n) ;  
                for(i = 0 ; i < n ; i++)  
                    s[i] = s[i]*s[i]*PI ;  
                low = 0 ;  
                high = s[n-1] ;  
                while( (high-low) > eqs )  
                {  
                    mid = (low+high)/2.0 ;  
                    if( solve(mid) )  
                    {  
                        low = mid  ;  
                        last = mid ;  
                    }  
                    else  
                        high = mid ;  
                }  
                printf("%.4lf
", last) ;  
            }  
        }  
        return 0;  
    }

 HDU2141

Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)

Total Submission(s): 1157 Accepted Submission(s): 373

Problem Description

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.

 

Input

There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.

 

Output

For each case, firstly you have to print the case number as the form "Case d:", then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print "YES", otherwise print "NO".

 

Sample Input

3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10

 

Sample Output

Case 1:
NO
YES
NO

 

Author

wangye

 

Source

HDU 2007-11 Programming Contest

 

Recommend

威士忌

 

一样的题目, 不多说了

代码:

#include<stdio.h>
#include<algorithm>
using namespace std;
#define K 505
int LN[K*K];
int BinarySearch(int LN[],int h,int t)/*二分查找*/
{
    int left,right,mid;
    left=0;
    right=h-1;
    mid=(left+right)/2;
    while(left<=right)
    {
        mid=(left+right)/2;
        if(LN[mid]==t)
          return 1;
        else if(LN[mid]>t)
          right=mid-1;
        else if(LN[mid]<t)
          left=mid+1;
    }
    return 0;
}
int main()
{
    int i,j,count=1,q;
    __int32 L[K],N[K],M[K],S,n,m,l;
    while(scanf("%d%d%d",&l,&n,&m)!=EOF)
    {
        int h=0;
        for(i=0;i<l;i++)
         scanf("%d",&L[i]);
        for(i=0;i<n;i++)
          scanf("%d",&N[i]);
        for(i=0;i<m;i++)
          scanf("%d",&M[i]);
        for(i=0;i<l;i++)
          for(j=0;j<n;j++)
           LN[h++]=L[i]+N[j];/*合并L和N*/
        sort(LN,LN+h); /*对LN数组排序*/
        scanf("%d",&S);
        printf("Case %d:
",count++);
        for(i=0;i<S;i++)
        {
            scanf("%d",&q);/*q即为题目中的x*/
            int p=0; /*p为标记，0为找不到，1为能找到*/
            for(j=0;j<m;j++)
            {
                int a=q-M[j]; /*因为L[i]+N[j]+M[k]==q，所以q-M[k]=LN[h]*/
                if(BinarySearch(LN,h,a)) /*在LN数组中查找到a*/
                {
                    printf("YES
");
                    p=1;
                    break;
                }
            }
            if(!p) /*找不到a*/
              printf("NO
");
        }
    }
    return 0;
}