poj 1701 Dissatisfying Lift （枚举）

╰半夏微凉° 2022-04-10 04:56 226阅读 0赞

Dissatisfying Lift

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;5102</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;1123</td> 
  </tr> 
 </tbody> 
</table>

Description

There's a building with M floors. The amounts of tenants of every floor are K1, K2, K3, ..., Km. One day all the tenants went home together and they took the same lift (suppose the lift was large enough). Because of some reason the lift could only stop on one floor and the tenants must go upstairs or downstairs to their houses. Every tenant went up N floors would make the dissatisfied degree rise N \* a + 0.5 \* N \* (N - 1) degrees, and every tenant went down N floors would make the dissatisfied degree rise N \* b + 0.5 \* N \* (N - 1) degrees. Your task is to tell which floor the lift should stop, in order to make the dissatisfied degree as low as possible.

Input

The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. The first line of each test contains M (1 <= M <= 10000), a and b (0 <= a, b <= 100). The second line contains K1, K2, K3, ..., Km (0 <= Ki <= 20, i = 1..M).

Output

For each test case, print a line containing a single integer, indicating which floor the lift should stop.

Sample Input

1
    5 3 2
    1 1 1 1 1

Sample Output

分析：则任意求任意两层不满衣服只差的最小值即为所要求的楼层。

dD(i+1,i)=Di+1 - Di;

花间得dD=(b+i)*sumk[i] - (a-i-1)*(sumk[M]-sumk[i])-c; #include <iostream>
    
    using namespace std;
    int T,M,a,b;
    int k[10010],sumk[10010];
    long long dD;   //dD=在第i+1层停靠的不满意度之和与在i层不满意度之和的差
    long long c;  //c=k[1]*1+k[2]*2+k[3]*3+....+k[m]*m;
    int ans;
    int main()
    {
        int i;
        cin>>T;
        while(T--)
        {
            cin>>M>>a>>b; //楼层数 上满意度 下满意度
            cin>>k[1];c=sumk[1]=k[1];
            for(i=2;i<=M;i++)
            {
                cin>>k[i];
                sumk[i] = sumk[i-1]+k[i];
                c+=k[i]*i;
            }
            //求解并输出
            ans=1;
            for(i=1;i<M;i++)
            {
                dD=(b+i)*sumk[i] - (a-i-1)*(sumk[M]-sumk[i])-c;
                if(dD<0) ans=i+1;
                else break;
            }
            cout<<ans<<endl;
        }
        return 0;
    }