学步园

During the next century certain regions on earth will experience severe water shortages. The old town of Uqbar has already started to prepare itself for the worst. Recently they created a network of pipes connecting the cisterns that distribute water in each
neighbourhood, making it easier to fill them at once from a single source of water. But in case of water shortage the cisterns above a certain level will be empty since the water will to the cisterns below. 

You have been asked to write a program to compute the level to which cisterns will be lled with a certain volume of water, given the dimensions and position of each cistern. To simplify we will neglect the volume of water in the pipes. 

Task 

Write a program which for each data set: 

reads the description of cisterns and the volume of water, 

computes the level to which the cisterns will be filled with the given amount of water, 

writes the result. 

The first line of the input contains the number of data sets k, 1 <= k <= 30. The data sets follow. 

The first line of each data set contains one integer n, the number of cisterns, 1 <= n <= 50 000. Each of the following n lines consists of 4 nonnegative integers, separated by single spaces: b, h, w, d - the base level of the cistern, its height, width and
depth in meters, respectively. The integers satisfy 0 <= b <= 10^6 and 1 <= h * w * d <= 40 000. The last line of the data set contains an integer V - the volume of water in cubic meters to be injected into the network. Integer V satisfies 1 <= V <= 2 * 10^9. 

The output should consist of exactly d lines, one line for each data set. 

Line i, 1 <= i <= d, should contain the level that the water will reach, in meters, rounded up to two fractional digits, or the word 'OVERFLOW', if the volume of water exceeds the total capacity of the cisterns. 

3
2
0 1 1 1
2 1 1 1
1
4
11 7 5 1
15 6 2 2
5 8 5 1
19 4 8 1
132
4
11 7 5 1
15 6 2 2
5 8 5 1
19 4 8 1
78

1.00
OVERFLOW
17.00

Central Europe 2001

题意：求给定的水，可以把连通器装到多少高度，若有水剩余则输出OVERFLOW

题解：二分高度，然后枚举计算其所需的水量就行了，注意控制好精度（一开始我竟然将double当int传。。弱啊）

#include<stdio.h>
#define eps 1e-4
struct cistern{
    double b,h,w,d;
}cc[50008];
int n;
double MIN(double x,double y){ return x<y?x:y; }
double cal(double mid)
{
    int i;
    double res=0;

    for(i=0;i<n;i++)
    {
        if(mid>cc[i].b)
        {
            res+=MIN(cc[i].h,mid-cc[i].b)*cc[i].w*cc[i].d;
        }
    }

    return res;
}
double solve(double all)
{
    double left=0,right=1000000000,mid;

    while(left+eps<right)
    {
        mid=(left+right)/2.0;
        if(cal(mid)<all) left=mid;
        else right=mid;
    }

    return left;
}
int main()
{
    int t,i;
    double all,overflow;

    //freopen("t.txt","r",stdin);
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(overflow=i=0;i<n;i++)
        {
            scanf("%lf%lf%lf%lf",&cc[i].b,&cc[i].h,&cc[i].w,&cc[i].d);
            overflow+=cc[i].h*cc[i].w*cc[i].d;
        }
        scanf("%lf",&all);
        if(overflow<all) printf("OVERFLOW\n");
        else printf("%.2lf\n",solve(all)+eps);
    }
}