poj 2976(二分)

题意：有一个式子是

给出n组a[i]和b[i]，去掉k组，让剩下的a[i]和b[i]带入式子计算出的最大值是多少。

题解：和poj3111的思路类似，只不过这个要四舍五入，注意强制转换。

#include <stdio.h>
#include <algorithm>
using namespace std;
const int N = 1005;
struct P {
    int a, b;
    double x;
}p[N];
int n, k;

bool cmp(P a, P b) {
    return a.x > b.x;
}

bool judge(double s) {
    for (int i = 0; i < n; i++)
        p[i].x = p[i].a * 1.0 - s * p[i].b; 
    sort(p, p + n, cmp);
    double temp = 0;
    for (int i = 0; i < k; i++)
        temp += p[i].x;
    return temp >= 0;
}

int main() {
    while (scanf("%d%d", &n, &k) == 2 && n + k) {
        k = n - k;
        double l = 0, r = 0;
        for (int i = 0; i < n; i++) {
            scanf("%d", &p[i].a);
            r += p[i].a;
        }
        for (int i = 0; i < n; i++)
            scanf("%d", &p[i].b);
        for (int i = 0; i < 100; i++) {
            double mid = (l + r) / 2;
            if (judge(mid))
                l = mid;
            else
                r = mid;
        }
        printf("%d\n", (int)(100 * (l + 0.005)));
    }
    return 0;
}