const double EPS = 1e-4;
const int MAXN = 200;
double c[MAXN], t[MAXN], r[MAXN];
int n, q, kase = 1;
double otp, inp;
bool check(double time, double all)
{
double cost = 0;
REP(i, n)
cost += max(0.0, time * (c[i] / t[i]) - r[i]);
return cost <= all;
}
int main()
{
// freopen("in.txt", "r", stdin);
while (~RII(n, q) && n)
{
otp = 0;
printf("Case %d:\n", kase++);
scanf("%lf", &inp);
REP(i, n)
{
scanf("%lf%lf%lf", &c[i], &t[i], &r[i]);
otp += c[i] / t[i];
}
printf("%.0f\n", ceil(otp / inp));
REP(i, q)
{
int t;
RI(t);
double l = 0, r = 100001;
while (l + EPS <= r)
{
double m = (l + r) / 2;
if (check(m, t * inp * m))
l = m;
else
r = m;
}
if (l > 100000)
puts("-1.000");
else
printf("%.3f\n", l);
}
}
return 0;
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原文:http://blog.csdn.net/wty__/article/details/25804609