UVa12230 - Crossing Rivers (数学期望)
2015-05-06 22:02
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题意:给你A和B的距离,以及途中n条河的信息,问你从A到B的期望 思路:我们单纯算过河时间的话,最快的是l/v,最慢的可能是3l/v,期间的时间是线性的,所以期望就是4l/2v=2l/v
#include<iostream> #include<cstring> #include<algorithm> #include<cstdio> int main() { int n; double p,l,v,d; int cas=1; while(scanf("%d%lf",&n,&d)!=EOF&&n+d){ while(n--){ scanf("%lf%lf%lf",&p,&l,&v); d=d-l+l*2/v; } printf("Case %d: %.3lf\n\n",cas++,d); } return 0; }
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