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; }