Codeforces Round #341 (Div. 2) C Wet Shark and Flowers - 期望与概率
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题目描述
- 标准题解
- 针对每一个
sharki ,算出 [l,r] 之间有多少个p的倍数,记为ai
ai=⌊rip⌋+⌊li−1p⌋ - 因为对于任意相邻的shark得到的money是独立的,可以针对pair(i,i+1) (注意1和n也是相邻的) 计算出他俩在整个过程中得到的money,记为
Gi Gi=2000∗(ai∗(rj−lj+1)+aj∗(ri−li+1)−ai∗aj)∗TS
化简之后,得到:
ans用double类型,边除边乘就不用担心溢出的问题了。
#include<cstdio>#define MAXN 100000double expt=0;int n,l[MAXN+10],r[MAXN+10],a[MAXN+10],p;int main(){ int t1,t2; scanf("%d%d",&n,&p); for(int i=1;i<=n;i++){ scanf("%d%d",&l[i],&r[i]); a[i]=r[i]/p-(l[i]-1)/p; } for(int i=1;i<n;i++){ t1=r[i+1]-l[i+1]+1,t2=r[i]-l[i]+1; expt+=(1.0*t1*t2-1.0*(t1-a[i+1])*(t2-a[i]))*2000.0/t1/t2; } t1=r[1]-l[1]+1,t2=r[n]-l[n]+1; expt+=(1.0*t1*t2-1.0*(t1-a[1])*(t2-a[n]))*2000.0/t1/t2; printf("%.7lf\n",expt);}
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