ZOJ problem Set-3640 help Me Escape

Main topic:

There are N road, choose the probability of each road is equal, the initial capacity value is f, each road through the difficulty value of CI, when the capacity value is greater than the difficulty of a road a B, can successfully flee, spend time ti, less than equals, can not escape, days plus a day, but the ability to increase the value of B.

Given the initial capability value, the expectation of a successful escape is obtained.

Analysis:

The probability of DP do less, the feeling is not very simple.

Set DP[J] Indicates the expected value of the escape when the capacity is J.

For each road I, when J>c[i], the successful Escape + (ti[i]*p), otherwise plus (+1+dp[j+c[j]]) *p;

From the back forward, to find out dp[f].

Precision Card is very strict, look at the following 2 code on the line

The code of WA

1#include <cstdio>2#include <cstring>3#include <cmath>4#include <algorithm>5#include <iostream>6 #defineINF 100000007 using namespacestd;8 intMain ()9 {Ten     intn,f; One     Doubledp[100000]; A     intc[100000],t[100000]; -      while(SCANF ("%d%d", &n,&f)! =EOF) -     { the         intsum=0, max=-INF; -          for(intI=0; i<n; i++) -         { -scanf"%d",&c[i]); +max=Max (max,c[i]); -             intTt= (int)((1.0+SQRT (5.0))/2.0*c[i]*c[i]); +t[i]=tt; Asum+=T[i]; at         } -         Doublep=1.0/N; -          for(inti=max+1; i<=2*max; i++) -Dp[i]= (Double) sum*p; -          for(intJ=max; j>=f; j--) -         { in             Doubletem=0.0; -              for(intI=0; i<n; i++) to             { +                 if(c[i]<j) -tem+=t[i]*p; the                 Else *tem+= (1+dp[j+c[i]]) *p; $             }Panax Notoginsengdp[j]=tem; -         } theprintf"%.3lf\n", Dp[f]); +     } A     return 0; the}

AC Code

1#include <cstdio>2#include <cstring>3#include <cmath>4#include <algorithm>5#include <iostream>6 #defineINF 100000007 using namespacestd;8 intMain ()9 {Ten     intn,f; One     Doubledp[100000]; A     intc[100000],t[100000]; -      while(SCANF ("%d%d", &n,&f)! =EOF) -     { the         Doublesum=0; -         intmax=-INF; -          for(intI=0; i<n;i++) -         { +scanf"%d",&c[i]); -max=Max (max,c[i]); +             intTt= (int)((1.0+SQRT (5.0))/2.0*c[i]*c[i]); At[i]=tt; atsum+=T[i]; -         } -         Doublep=1.0/N; -          for(inti=max+1; i<=2*max;i++) -dp[i]=sum*p; -          for(intj=max;j>=f;j--) in         { -             Doubletem=0.0; to              for(intI=0; i<n;i++) +             { -                 if(c[i]<j) thetem+=t[i]*p; *                 Else $tem+= (1+dp[j+c[i]]) *p;Panax Notoginseng             } -dp[j]=tem; the         } +printf"%.3lf\n", Dp[f]); A     } the     return 0; +}

ZOJ problem Set-3640 help Me Escape