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  • 题解

    userId_undefined

    法兰西玫瑰

    倔强青铜
    20阅读
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  • Jt

    #include <iostream> using namespace std; double w, l; int n; double a[1100], b[1100], f[1100]; int main() { cin >> w >> l >> n; for (int i = 1; i <= n; i++) { cin >> a[i] >> b[i]; f[i] = 99999999999999999; } f[1] = l / b[1] * 60; f[0] = 0; for (int i = 2; i <= n; i++) { double weight = 0; double ms = b[i]; for (int j = i; j >= 1; j--) { weight += a[j]; if (weight > w) break; if (b[j] < ms) ms = b[j]; f[i] = min(f[i], f[j - 1] + l / ms * 60); } } printf("%.1f", f[n]); return 0; }

    userId_undefined

    马智辉

    倔强青铜
    3阅读
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  • Jt

    #include <iostream> using namespace std; double w, l; int n; double a[1100], b[1100], f[1100]; int main() { cin >> w >> l >> n; for (int i = 1; i <= n; i++) { cin >> a[i] >> b[i]; f[i] = 99999999999999999; } f[1] = l / b[1] * 60; f[0] = 0; for (int i = 2; i <= n; i++) { double weight = 0; double ms = b[i]; for (int j = i; j >= 1; j--) { weight += a[j]; if (weight > w) break; if (b[j] < ms) ms = b[j]; f[i] = min(f[i], f[j - 1] + l / ms * 60); } } printf("%.1f", f[n]); return 0; }

    userId_undefined

    马智辉

    倔强青铜
    2阅读
    0回复
    0点赞
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