HDU 1551 Cable master 二分-优快云博客

本文探讨了如何使用二分查找算法解决将N条电缆截成相同长度M段的问题，提供了详细的题解步骤及代码实现，旨在帮助读者理解并掌握此算法的应用。

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题意：把N条电缆截成相同的M段，问截成的最大长度是多少？

思路：二分截的长度即可。（我不会告诉你我的程序样例都过不了）

http://acm.hdu.edu.cn/showproblem.php?pid=1551

/*********************************************
    Problem : HDU 1551
    Author  : NMfloat
    InkTime (c) NM . All Rights Reserved .
********************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++)
#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --)
#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)
#define cls(a,x)   memset(a,x,sizeof(a))
#define eps 1e-8

using namespace std;

const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5+5;
const int MAXE = 2e5+5;

typedef long long LL;
typedef unsigned long long ULL;

int T,n,m,k;

int fx[] = {0,1,-1,0,0};
int fy[] = {0,0,0,-1,1};
double A[10005] ;
double rights;
double sum;

void input() {
    double tmp;
    rights = 0;
    rep(i,1,n) {
        scanf("%lf",&A[i]);
        rights += A[i];
    }
}

int calc(double mid) {
    int count = 0;
    rep(i,1,n) {
        count += (int)( A[i] / mid );
    }
    return count;
}

void solve() {
    double left = 0;
    double right = rights;
    double mid ;
    while(fabs(right-left)>eps) {
        mid = (left + right) / 2;
        if(calc(mid) < m) right = mid;
        else left = mid;
    }
    printf("%.2lf\n",mid);
}

int main(void) {
    //freopen("a.in","r",stdin);
    while(scanf("%d %d",&n,&m),n+m) {
        input();
        solve();
    }
    return 0;
}