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 = ;

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

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

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

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

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

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