題意:給你n條電纜,分成k截,每截最長是多少?
找一個長度,判斷它能產生多少截繩子。沒錯,二分法。
需要注意的是,輸出的時候要小心四捨五入是不行的。(長度超長,不會產生相應的繩子)
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 10000 + 5;
const double eps = 1e-4;
double cable[maxn];
int n, k;
int getNum(double Len)
{
int sum = 0;
for(int i = 0; i < n; i++)
sum += cable[i] / Len;
return sum;
}
int main()
{
while(scanf("%d%d", &n, &k) == 2 && (n || k))
{
double L = 0, R = 0;
for(int i = 0; i < n; i++)
{
scanf("%lf", &cable[i]);
R = max(R, cable[i]);
}
while(R - L >= eps)
{
double mid = (R + L) / 2;
if(getNum(mid) < k) R = mid;//當產生相同截數時,我們要知道,應該使答案接近上限,應該提升下限。
else L = mid;
//cout<<" "<<L<<" "<<R<<endl;
}
int t = R * 100;
double ans = t / 100.0;
printf("%.2lf\n", ans);
}
return 0;
}