poj1064 Cable master

传送门

Time cost:35min

题意:给n根绳子 问切成k段的最大长度

妥妥的二分 能切成长的一定能切成短的

所以就O(lgV)二分 * O(n)判断能切成多少段

如果能切不少于k就L=m 否则R=m

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define rep(i,a,n) for(int i = a;i <= n;++i)
#define per(i,n,a) for(int i = n;i >= a;--i)
#define ms(a,b) memset(a,b,sizeof a)
#define eps 1e-5
using namespace std;
typedef double D;
int read() {
    int as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') fu = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    as = as * 10 + c - '0';
    c = getchar();
    }
    return as * fu;
}
//head
const int N = 10006;
int n,k;
D a[N],maxx;
D L,R;
bool check(D l) {
    int ans = 0;
    rep(i,1,n) ans += int(a[i] / l);
    return ans >= k;
}

int main() {
    n = read();
    k = read();
    rep(i,1,n) scanf("%lf",&a[i]),maxx = max(maxx,a[i]);
    L = 0,R = maxx;
    while(R - L > eps) {
    D m = (L + R) / 2.000;
    check(m) ? L = m : R = m;
    }
    printf("%.2lf
",floor(R*100)/100);
    return 0;
}