Codeforces 185B（数学结论）

要点

• 有不等式\(x^ay^bz^c=a^ab^bc^c(\frac{x}{a})^a(\frac{y}{b})^b(\frac{z}{c})^c<=a^ab^bc^c(\frac{x+y+z}{a+b+c})^{a+b+c}\)，而本题中又有\(x+y+z<=S\)，故可直接获得。

#include 

int S, a, b, c;

int main() {
    scanf("%d%d%d%d", &S, &a, &b, &c);
    if (a + b + c == 0) {
        return !printf("0 0 0\n");
    }
    double x = (double)a / (a + b + c) * S;
    double y = (double)b / (a + b + c) * S;
    double z = (double)c / (a + b + c) * S;
    printf("%.20lf %.20lf %.20lf\n", x, y, z);
}

转载于:https://www.cnblogs.com/AlphaWA/p/10921479.html