二分搜索 POJ 1064 Cable master

 

题目传送门

 1 /*  2  题意：n条绳子问切割k条长度相等的最长长度  3  二分搜索：搜索长度，判断能否有k条长度相等的绳子  4 */  5 #include <cstdio>  6 #include <algorithm>  7 #include <cstring>  8 #include <cmath>  9 using namespace std; 10 11 const int MAXN = 1e4 + 10; 12 const int INF = 0x3f3f3f3f; 13 double w[MAXN]; 14 int n, k; 15 16 int check(double len) { 17 int ret = 0; 18 for (int i=1; i<=n; ++i) { 19 ret += (int) (w[i] / len); 20  } 21 return ret; 22 } 23 24 int main(void) { //POJ 1064 Cable master 25 //freopen ("POJ_1064.in", "r", stdin); 26 27 while (scanf ("%d%d", &n, &k) == 2) { 28 for (int i=1; i<=n; ++i) { 29 scanf ("%lf", &w[i]); 30  } 31 double l = 0, r = 1e9; 32 for (int i=1; i<=100; ++i) { 33 double mid = (l + r) / 2; 34 if (check (mid) >= k) l = mid; 35 else r = mid; 36  } 37 printf ("%.2f\n", floor (l * 100) / 100); 38  } 39 40 return 0; 41 }