铁人系列 （1） uva 10385

uva  10385

列出n-1个一元方程，对应成单峰函数，所以用三分求解即可。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 30;

int N;
double L, vr[maxn], vk[maxn];

void init () {
    for (int i = 1; i <= N; i++) {
        vr[i] = 1/vr[i] - 1/vk[i];
        vk[i] = L/vk[i];
    }

    for (int i = 1; i < N; i++) {
        vr[i] -= vr[N];
        vk[i] -= vk[N];
    }
}

double f (double x) {
    double ret = vr[1] * x + vk[1];
    for (int i = 2; i < N; i++)
        ret = min(ret, vr[i] * x + vk[i]);
    return ret;
}

double tsearch (double l, double r) {
    for (int i = 0; i < 100; i++) {
        double p = l + (r - l) / 3;
        double q = r - (r - l) / 3;
        if (f(p) > f(q))
            r = q;
        else
            l = p;
    }
    return l;
}

int main () {
    while (scanf("%lf%d", &L, &N) == 2) {
        for (int i = 1; i <= N; i++)
            scanf("%lf%lf", &vr[i], &vk[i]);
        init();
        double x = tsearch(0, L);
        if (f(x) < 0)
            printf("The cheater cannot win.
");
        else
            printf("The cheater can win by %.0lf seconds with r = %.2lfkm and k = %.2lfkm.
", f(x) * 3600, x, L-x);
    }
    return 0;
}