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Time cost:35min

題意:給n根繩子 問切成k段的最大長度

妥妥的二分 能切成長的一定能切成短的

所以就O(lgV)二分 * O(n)判斷能切成多少段

如果能切不少於k就L=m 否則R=m

Code:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cmath>
 5 #define rep(i,a,n) for(int i = a;i <= n;++i)
 6 #define per(i,n,a) for(int i = n;i >= a;--i)
 7
 
 #define ms(a,b) memset(a,b,sizeof a)
 8 #define eps 1e-5
 9 using namespace std;
10 typedef double D;
11 int read() {
12     int as = 0,fu = 1;
13     char c = getchar();
14     while(c < '0' || c > '9') {
15     if(c == '-') fu = -1;
16     c = getchar();
17     }
18     while(c >= '0' && c <= '
 
9') {
19     as = as * 10 + c - '0';
20     c = getchar();
21     }
22     return as * fu;
23 }
24 //head
25 const int N = 10006;
26 int n,k;
27 D a[N],maxx;
28 D L,R;
29 bool check(D l) {
30     int ans = 0;
31     rep(i,1,n) ans += int(a[i] / l);
32     return ans >= k;
33 }
34 
35 int main() {
36     n = read();
 
37     k = read();
38     rep(i,1,n) scanf("%lf",&a[i]),maxx = max(maxx,a[i]);
39     L = 0,R = maxx;
40     while(R - L > eps) {
41     D m = (L + R) / 2.000;
42     check(m) ? L = m : R = m;
43     }
44     printf("%.2lf\n",floor(R*100)/100);
45     return 0;
46 }