Codeforces Round #284 (Div. 2) D. Name That Tune [概率dp]-白红宇

Codeforces Round #284 (Div. 2) D. Name That Tune [概率dp]

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发布时间：2019-06-30

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It turns out that you are a great fan of rock band AC/PE. Peter learned that and started the following game: he plays the first song of the list of n songs of the group, and you have to find out the name of the song. After you tell the song name, Peter immediately plays the following song in order, and so on.

The i-th song of AC/PE has its recognizability p_i. This means that if the song has not yet been recognized by you, you listen to it for exactly one more second and with probability of p_i percent you recognize it and tell it's name. Otherwise you continue listening it. Note that you can only try to guess it only when it is integer number of seconds after the moment the song starts playing.

In all AC/PE songs the first words of chorus are the same as the title, so when you've heard the first t_i seconds of i-th song and its chorus starts, you immediately guess its name for sure.

For example, in the song Highway To Red the chorus sounds pretty late, but the song has high recognizability. In the song Back In Blue, on the other hand, the words from the title sound close to the beginning of the song, but it's hard to name it before hearing those words. You can name both of these songs during a few more first seconds.

Determine the expected number songs of you will recognize if the game lasts for exactly T seconds (i. e. you can make the last guess on the second T, after that the game stops).

If all songs are recognized faster than in T seconds, the game stops after the last song is recognized.

Input

The first line of the input contains numbers n and T (1 ≤ n ≤ 5000, 1 ≤ T ≤ 5000), separated by a space. Next n lines contain pairs of numbers p_i and t_i (0 ≤ p_i ≤ 100, 1 ≤ t_i ≤ T). The songs are given in the same order as in Petya's list.

Output

Output a single number — the expected number of the number of songs you will recognize in T seconds. Your answer will be considered correct if its absolute or relative error does not exceed 10^- 6.

Sample test(s)

Input

2 2 50 2 10 1

Output

1.500000000

Input

2 2 0 2 100 2

Output

1.000000000

Input

3 3 50 3 50 2 25 2

Output

1.687500000

Input

2 2 0 2 0 2

Output

1.000000000

感觉集齐了各种坑，萌萌哒~ 以后只能默默地给田神拎包了

2014-12-28 07:06:00	GNU C++	Accepted	483 ms	196200 KB
2014-12-28 07:01:48	GNU C++	Runtime error on test 36	93 ms	196100 KB
2014-12-28 06:54:26	GNU C++	Runtime error on test 36	109 ms	196100 KB
2014-12-28 06:49:32	GNU C++	Runtime error on test 36	93 ms	196200 KB
2014-12-28 06:46:58	GNU C++	Runtime error on test 36	93 ms	196100 KB
2014-12-28 06:44:55	GNU C++	Wrong answer on test 3	93 ms	196100 KB
2014-12-28 06:40:00	GNU C++	Wrong answer on test 3	62 ms	196100 KB
2014-12-28 06:36:34	GNU C++	Wrong answer on test 3	109 ms	196100 KB
2014-12-28 06:25:16	GNU C++	Wrong answer on test 7	62 ms	196200 KB
2014-12-28 06:20:57	GNU C++	Wrong answer on test 7	124 ms	196100 KB
2014-12-26 04:14:38	GNU C++	Time limit exceeded on test 13	1000 ms	196200 KB
2014-12-26 04:09:52	GNU C++	Time limit exceeded on test 13	1000 ms	196200 KB
2014-12-25 16:29:08	GNU C++	Wrong answer on test 6	62 ms	196200 KB
2014-12-25 16:25:22	GNU C++	Memory limit exceeded on test 1	108 ms	262100 KB
2014-12-25 16:23:15	GNU C++	Memory limit exceeded on test 1	93 ms	262100 KB

1 #include
     
       2 #include
      
        3 #include
       
         4 #include
        
          5 #include
         
           6 #include
           7 #include
           
             8 #include
            
              9 #include
             
              10 #include
              
               11 #define eps 1e-312 #define ll long long13 #define N 500514 #define M 1000515 #define mod 200716 17 using namespace std;18 19 int n,T;20 double dp[N][N];21 double ans;22 double tot;23 double p[N];24 int t[N];25 //double q[N][N];26 27 void ini()28 {29 ans=0;tot=0;30 T++;31 memset(dp,0,sizeof(dp));32 // memset(q,0,sizeof(q));33 int i;34 for(i=1;i<=n;i++){35 scanf("%lf%d",&p[i],&t[i]);36 p[i]/=100;37 }38 dp[1][0]=1.0;39 //for(i=1;i<=n;i++){40 // q[i][1]=p[i];41 // q[i][ t[i] ]=1;42 // for(j=2;j<=t[i]-1;j++){43 // q[i][j]=q[i][j-1]*(1.0-p[i]);44 // }45 // }46 }47 48 void solve()49 {50 int i,j;51 int tmin,tmax;52 tmin=1,tmax=1;53 //double tmp;54 for(i=1;i<=min(n,T);i++){55 tmin++;tmax=min(T,tmax+t[i]);56 // tmp=pow(1-p[i],t[i]-1);57 for(j=tmin;j

转载于:https://www.cnblogs.com/njczy2010/p/4189806.html

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