poj 1064 Cable master【浮点型二分查找】
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 29554 | Accepted: 6247 |
Description
To buy network cables, the Judging Committee has contacted a local network solutions provider with a request to sell for them a specified number of cables with equal lengths. The Judging Committee wants the cables to be as long as possible to sit contestants as far from each other as possible.
The Cable Master of the company was assigned to the task. He knows the length of each cable in the stock up to a centimeter,and he can cut them with a centimeter precision being told the length of the pieces he must cut. However, this time, the length is not known and the Cable Master is completely puzzled.
You are to help the Cable Master, by writing a program that will determine the maximal possible length of a cable piece that can be cut from the cables in the stock, to get the specified number of pieces.
Input
Output
If it is not possible to cut the requested number of pieces each one being at least one centimeter long, then the output file must contain the single number "0.00" (without quotes).
Sample Input
4 11
8.02
7.43
4.57
5.39
Sample Output
2.00
1、最后的r输出是要用%f
2、刚开始一直错,看了别人的题解,发现原来并不一定要把所有的电缆都用上如数据
2 2
1.02
2.33
答案为1.16
3、还有就是要注意最后的结果并不是四舍五入,而是直接取小数点后两位
#include<stdio.h>
#include<string.h>
#define MAX 10100
#define maxn(x,y)(x>y?x:y)
double a[MAX];
int n,m;
int num(double x)//计算在x长度时最多可以有多少段这样做
{ //就解决了不一定全部电缆都要用的问题
int i,sum=0;
for(i=1;i<=n;i++)
{
sum+=(int)(a[i]/x);
}
return sum;
}
int main()
{
int i;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(a,0,sizeof(a));
double max=0,suma=0;
for(i=1;i<=n;i++)
{
scanf("%lf",&a[i]);
max=maxn(max,a[i]);
suma+=a[i];
}
if(suma/m<0.01)
printf("0.00\n");
else
{
double l=0.01,r=max,mid=0;
while(r - l >1e-8)
{
mid=(l+r)/2.0;
if(num(mid)>=m)
l=mid;
else
r=mid;
}
int tem=r*100;
double ans=tem*0.01;
printf("%.2f\n",ans);
}
}
return 0;
}
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