暴力解法+一点点分析

a+b=5 b+c=3 c+c=2 又隐含地有a，b，c>0 其实可以得出c=1，但装装样子，写个循环 显然有：a<=5,b<=3

    for j in range(0,4):     # 0<=十位数<=3
        for k in range(0,3): # 0<=个位数<=2
            if i*100+j*10+k+j*100+k*11==532:
                print(str(i),str(j),str(k))