题目写的比较难懂，题目大意：给你一个n，m代表有n首歌，在m秒内进行猜歌。让你求出m秒过后你能猜出来的歌曲的数量的期望。接下来n行，每行有两个数字第一个pi代表在歌曲结束前你猜出来的概率，第二个ti代表这首歌最多播放ti秒，在t秒之后你百分比之百能猜出来歌曲。最后输出m秒之后猜出来的歌曲的期望。
我们设dp[i][j]表示枚举到第i首歌，所用时间恰好为j的期望。
dp[i][j] 可以由状态dp[i-1][j-ti],dp[i-1][j-ti+1],......,dp[i-1][j-1]转移过来。
dp[i][j] = dp[i-1][j-ti]*(1-pi)^(ti-1)+dp[i-1][j-ti+1]*(1-pi)^(ti-2)*p+dp[i-1][j-ti+2]*(1-pi)^(ti-3)*p+......+dp[i-1][j-1]*p
需要注意的是dp[i][j-ti]此时第i首歌曲一定会猜中。所以dp[i][j] += dp[i-1][j-ti]*pow((1-pi), ti);
还有就是要减去之前无用的状态，dp[i-1][j-ti-1]的状态是推不到dp[i][j]的所以要减去。
设ans=dp[i-1][j-ti+1]*(1-pi)^(ti-2)+dp[i-1][j-ti+2]*(1-pi)^(ti-3)+......+dp[i-1][j-1]，这样从j推向j+1时只要ans=ans*(1-pi)+dp[i-1][j]
B. Name That Tune
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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 pi.
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 pi 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 ti 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 pi and ti (0?≤?pi?≤?100, 1?≤?ti?≤?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
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 1000000007
#define Read() freopen("autocomplete.in","r",stdin)
#define Write() freopen("autocomplete.out","w",stdout)
#define Cin() ios::sync_with_stdio(false)
using namespace std;
inline int read()
{
char ch;
bool flag = false;
int a = 0;
while(!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')));
if(ch != '-')
{
a *= 10;
a += ch - '0';
}
else
{
flag = true;
}
while(((ch = getchar()) >= '0') && (ch <= '9'))
{
a *= 10;
a += ch - '0';
}
if(flag)
{
a = -a;
}
return a;
}
void write(int a)
{
if(a < 0)
{
putchar('-');
a = -a;
}
if(a >= 10)
{
write(a / 10);
}
putchar(a % 10 + '0');
}
const int maxn = 5050;
struct node
{
double p;
int ti;
}f[maxn];
double dp[maxn][maxn];
int main()
{
int n, m;
while(cin >>n>>m)
{
for(int i = 1; i <= n; i++)
{
scanf("%lf %d", &f[i].p, &f[i].ti);
f[i].p /= 100.0;
}
for(int i = 0; i <= m; i++) dp[0][i] = 0;
dp[0][0] = 1;
double sum = 0.0;
for(int i = 1; i <= n; i++)
{
double ans = 0;
double q = pow(1-f[i].p, 1.0*f[i].ti);
for(int j = 0; j <= m; j++)
{
if(j-1 >= 0) ans += dp[i-1][j-1];
if((j-f[i].ti-1) >= 0) ans -= dp[i-1][j-f[i].ti-1]*q;
dp[i][j] = ans*f[i].p;
if(j-f[i].ti >= 0) dp[i][j] += dp[i-1][j-f[i].ti]*q;
ans *= (1.0-f[i].p);
sum += dp[i][j];
}
}
printf("%.10lf\n", sum);
}
return 0;
}