int area 存在 lan freopen 最小 numbers esp between

The swimming area of Berhattan‘s city beach is marked out with n buoys. The buoys form a straight line. When the buoys were being put into the water, nobody cared to observe the same distance between each pair of adjacent buoys.

Now the beach keeper wants the distance between any two adjacent buoys to be the same. He plans to shift some or all of the buoys without changing their respective order. To facilitate the task, he wants the total length of all shifts to be as small as possible.

Given coordinates of the buoys, you should find the minimum possible length of all shifts, as well as new coordinates of the buoys.

Input

The first line of input contains a single integer n (2?≤?n?≤?400), n — the number of buoys. The second line contains buoys‘ integer coordinates x₁,?x₂,?...,?x_n (?-?10000?≤?x

_i?≤?10000). No two given buoys will share the same place. The coordinates are given in strictly increasing order.

Output

To the first line print a real number t — the minimum possible total length of required shifts. Output this value with at least 4 digits after the decimal point.

To the second line print n

numbers — new coordinates of the buoys. The new coordinates should be printed in strictly increasing order with at least 7 digits after the decimal point. If there are several optimal ways to shift the buoys, you may output any of them.

Example Input

4
-2 2 6 9

Output

1.0000
-2.0000000000 1.6666666667 5.3333333333 9.0000000000

Note

All buoys are located on the Ox axis. You may move buoys only along the Ox axis.

題意 ： 在 X 軸上有一串浮標，移動一些浮標，讓他們之間的距離相等，問最小移動的總距離之和是多少，並輸出此時的浮標位置

思路 ： 對於兩個浮標間的距離 X ，當X過大或過小的時候都會使答案很大，因此中間存在一個最小值，這裏三分就可以，然後既然要使距離最小，那我們就先確保一個浮標不動去移動其他的，這裏枚舉就可以。

代碼示例：

#define ll long long
const int maxn = 1e6+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

double pre[405];
double ans[405];
double sum = 1.0*inf, lenn;
int n, pos;

double fun(double len, int base){
    double x = pre[base];
    double s = 0;
    for(int i = base-1; i >= 1; i--){
        x -= len;
        s += fabs(x-pre[i]);
    }
    x = pre[base];
    for(int i = base+1; i <= n; i++){
        x += len;
        s += fabs(x-pre[i]);
    }
    if (s < sum) {
        sum = s;
        lenn = len; pos = base;
    } 
    return s;
}

int main() {
   freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    
    cin >> n;
    for(int i = 1; i <= n; i++){
        scanf("%lf", &pre[i]);
    }
    
    for(int i = 1; i <= n; i++){
        double l = 0, r = 200000;
        for(int j = 1; j <= 100; j++){
            double lm = l + (r-l)/3;
            double rm = r - (r-l)/3;
            double lf = fun(lm, i);
            double rf = fun(rm, i);
            if (lf > rf) l = lm;
            else r = rm; 
        }   
    }
    ans[pos] = pre[pos];
    for(int i = pos+1; i <= n; i++) ans[i] = ans[i-1] + lenn;
    for(int i = pos-1; i >= 1; i--) ans[i] = ans[i+1] - lenn; 
    printf("%.4f\n", sum);
    for(int i = 1; i <= n; i++) printf("%.10f%c", ans[i], i==n?‘\n‘:‘ ‘);
    return 0;
}

枚舉 + 三分 （遊標）