問題 I: Kid and Ants

時間限制:?1 Sec??內存限制:?128 MB
提交:?42??解決:?33
[提交][狀態][討論版]

題目描述

Kid?likes?interest?question,although?he?don’t?like?ants?so?much.

Assume?there?is?a?infinite?long?stick?whose?direction?is?from?East?to?West.?In?addition,?there?are?n

ants?on?the?stick?and?their?original?orientation?is?arbitrary(East?or?West).Then?from?a?moment,all

the?ants?will?climb?on?the?stick?at?the?same?speed?toward?their?original?orientation.Once?two?ants

touch?each?other,?they?will?change?their?orientation?and?go?up?climbing.(May?be?some?ants?will

never?meet?other?ants).So?now?Kid’s?task?is?to?calculate?the?mathematical?expectation?of?the?meets

between?ants.

?

輸入

There?are?multiple?cases.

For?each?case,there?is?a?integer?n.(1<=n<=10^9)

?

輸出

For?each?case,just?print?the?mathematical?expectation(retain?three?decimals)

?

樣例輸入

2 10

樣例輸出

0.250 11.250

提示

?

When?n=2,there?4?cases(E?indicates?the?orientation?of?the?ant?is?East?and?W?is?West)

(left?is?East?and?right?is?West)

{E,E},{E,W},{W,E},{W,W}

And?the?number?of?meets?is?0,0,1,0,?so?the?mathematical?expectation?is?(0+0+1+0)/4=0.250.

?

隨便取兩只螞蟻,方案數n*(n-1)/2,對這兩只螞蟻有四種狀態,一種碰撞 #include <cstdio> using namespace std; int main(){int n;while(scanf("%lf",&n)==1&&n){printf("%.3f\n",(double)n*(n-1)/8.0); }return 0; }

　　

轉載于:https://www.cnblogs.com/xuesu/p/4264332.html

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