think in a 2nd way
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worm speed is < the series of : .01 * ( 1 + 1/n + 1/n*(n-1) + 1/ n*(n-1)*(n-2) ... )
sum : .01 * ( n + 1+1/2 + 1/3 ... + 1/2 + 1/6 + 1/12 ... + 1/6+1/24+1/60 ... ... )
< .01 * ( n + ln(n) + (1-1/n) + (1-1/n)/3 + (1-1/n)/4 ... )
< .01 * ( n + ln(n) + (1-1/n) * ln(n) )
< n

unless worm speed > 1. else will never catch up.
.
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这里好冷清,再出两个数学/物理题,活跃一下气氛 大树下   (455 bytes , 949reads )
我的解法-第二题 大树下   (1061 bytes , 412reads )
我的解法-第一题 大树下   (1157 bytes , 368reads )
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... 大树下   (60 bytes , 319reads )
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老兄你在乱猜是吧 大树下   (16 bytes , 299reads )
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think in a 2nd way 1300cc   (343 bytes , 506reads )
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while. we need formula to prove that. think e.g the worm speed is a function 1300cc   (135 bytes , 277reads )
speed converge to v1+v2 大树下   (880 bytes , 401reads )
i solve the integration wrongly 1300cc   (125 bytes , 227reads )
actually 1300cc   (483 bytes , 261reads )
approaching v1 + v2. 大树下   (0 bytes , 325reads )
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很久以前都做过了.. hash   (0 bytes , 268reads )
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