• 等差数列


    B. Trees in a Row
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    The Queen of England has n trees growing in a row in her garden. At that, the i-th (1 ≤ i ≤ n) tree from the left has height ai meters. Today the Queen decided to update the scenery of her garden. She wants the trees' heights to meet the condition: for all i (1 ≤ i < n),ai + 1 - ai = k, where k is the number the Queen chose.

    Unfortunately, the royal gardener is not a machine and he cannot fulfill the desire of the Queen instantly! In one minute, the gardener can either decrease the height of a tree to any positive integer height or increase the height of a tree to any positive integer height. How should the royal gardener act to fulfill a whim of Her Majesty in the minimum number of minutes?

    Input

    The first line contains two space-separated integers: nk (1 ≤ n, k ≤ 1000). The second line contains n space-separated integersa1, a2, ..., an (1 ≤ ai ≤ 1000) — the heights of the trees in the row.

    Output

    In the first line print a single integer p — the minimum number of minutes the gardener needs. In the next p lines print the description of his actions.

    If the gardener needs to increase the height of the j-th (1 ≤ j ≤ n) tree from the left by x (x ≥ 1) meters, then print in the corresponding line "+ j x". If the gardener needs to decrease the height of the j-th (1 ≤ j ≤ n) tree from the left by x (x ≥ 1) meters, print on the corresponding line "- j x".

    If there are multiple ways to make a row of trees beautiful in the minimum number of actions, you are allowed to print any of them.

    Sample test(s)
    input
    4 1
    1 2 1 5
    output
    2
    + 3 2
    - 4 1
    input
    4 1
    1 2 3 4
    output
    0


    解题方法:暴力求解,将分别从1,2,3,4,5……开始的等差数列与所给数据比较
    #include <iostream>
    #include <cstring>
    using namespace std;
    int a[1005],b[1005];
    
    int main()
    {
        int n,k;
        int i,j;
        while(cin>>n>>k)
        {
            memset(b,0,sizeof(b));
            for(i=0;i<n;i++)
            cin>>a[i];
            for(i=1;i<1005;i++)
            {
                for(j=0;j<n;j++)
                {
                    if(a[j]==i+j*k)
                    b[i]++;
                }
            }
            int y=b[0];
            int f=0;
            for(i=1;i<1005;i++)
            if(y<b[i])
            {
                y=b[i];
                f=i;
            }
            cout<<n-y<<endl;
            for(i=0;i<n;i++)
            {
                if(a[i]<f+i*k)
                cout<<"+ "<<i+1<<" "<<f+i*k-a[i]<<endl;
                if(a[i]>f+i*k)
                cout<<"- "<<i+1<<" "<<a[i]-f-i*k<<endl;
            }
        }
        return 0;
    }

    错误思想:将输入数据分别减去(i-1)*k,如果输入数据是等差数列,这时每个数应该相等(实际输入不是等差数列),从中找出相同数最多的数,其它数则为应当增减的数,此方法存在漏洞:若输入数据为5个,d=1,输入1 1 2 3 4,则处理后为0 1 2 3 4,而题中要求数据大于0,故错误!其代码如下:

    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <cmath>
    
    using namespace std;
    int a[1005],b[10000000];
    
    int main()
    {
        int n,k;
        int i,j,t,x,maxn,minn;
        while(cin>>n>>k)
        {
            minn=99999999;
            maxn=-99999999;
            memset(b,0,sizeof(b));
            for(i=0;i<n;i++)
            {
                scanf("%d",&x);
                a[i]=x-i*k;
                if(a[i]>maxn) maxn=a[i];
                if(a[i]<minn) minn=a[i];
            }
            t=0;
            for(i=maxn;i>=minn;i--)
            {
                for(j=0;j<n;j++)
                {
                    if(a[j]==i)
                    b[t]++;
                }
                t++;
            }
    
            int y=b[0];
            x=maxn-0;
            for(i=1;i<t;i++)
            {
                if(y<b[i])
                {
                    y=b[i];
                    x=maxn-i;
                }
            }
            cout<<n-y<<endl;
            for(i=0;i<n;i++)
            {
                if(a[i]>x)
                cout<<"- "<<i+1<<" "<<a[i]-x<<endl;
                if(a[i]<x)
                cout<<"+ "<<i+1<<" "<<x-a[i]<<endl;
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/chen9510/p/4943834.html
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