这可以说是一道数论（？ 首先可以发现$x$的取值为$[0,min(a,S/n)]$,$$$y$$$的取值为$$$S-xn$$$ $$$x$$$的取值最大为$$$min(a,S/n)$$$，得$$$y$$$的取值为$$$S-min(a,S/n)n$$$ 只要判断$y$是否小于等于$b$就好了
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