• POJ 2976 Dropping tests (二分+贪心)


    题意:给定 n 个分数,然后让你去年 m 个分数,使得把剩下的所有的分子和分母都相加的分数最大。

    析:这个题并不是分子越大最后结果就越大,也不是整个分数越大,最后结果就越大的,我们可以反过来理解,要去掉 m 个分数,那么就是要选 n-m个分数,

    那么就是 sigma(分子) / sigma(分母) 尽量大,那么最大是多大啊?这个我们可以通过二分来解决,也就是sigma(分子) / sigma(分母) >= x,

    因为分子和分母都是正数,所以可以得到 sigma(分子) - sigma(分母)* x  >= 0,也就是 sigma(分子 - x * 分母) >= 0(前n-m项),我们就可以按这个进行排序,

    看看前 n-m 项成不成立。

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    #include <sstream>
    #define debug() puts("++++");
    #define gcd(a, b) __gcd(a, b)
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 1e3 + 10;
    const int mod = 1e9 + 7;
    const int dr[] = {-1, 0, 1, 0};
    const int dc[] = {0, 1, 0, -1};
    const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline bool is_in(int r, int c){
      return r >= 0 && r < n && c >= 0 && c < m;
    }
    double c;
    
    struct Node{
      double a, b;
      bool operator < (const Node &p) const{
        return a - b * c > p.a - p.b * c;
      }
    };
    Node a[maxn];
    
    bool judge(){
      sort(a, a + n);
      double ans = 0;
      for(int i = 0; i < n-m; ++i)  ans += a[i].a- a[i].b * c;
      return ans >= 0.0;
    }
    
    double solve(){
      double l = 0, r = INF;
      for(int i = 0; i < 100; ++i){
        c = (l+r) / 2;
        if(judge())  l = c;
        else r = c;
      }
      return l;
    }
    
    int main(){
      while(scanf("%d %d", &n, &m) == 2 && n+m){
        for(int i = 0; i < n ; ++i)  scanf("%lf", &a[i].a);
        for(int i = 0; i < n ; ++i)  scanf("%lf", &a[i].b);
        printf("%.f
    ", solve()*100);
      }
      return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/6591076.html
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