[HDU 4870] Rating [DP], hdu4870rating

[HDU 4870] Rating [DP], hdu4870rating

A person registers two accounts, and the initial rating is 0. He plays the game with the low score each time. After winning the game, he adds 50 points and deducts 100 points. The winning rate is p, he will hit the game until he scored 1000 points on the first day and asked about the expectations of the game.

Question: because each time we increase or decrease the score is a multiple of 50, we can compress it into adding one point for each win and reducing the score by 2. According to the meaning, we can easily see that each score change is the smallest score.

Therefore, we define the State ans [I] [j] to indicate that the score from the initial state to the two numbers is I, j's expectation. We can know that the score change of the two numbers is always [I, I]-> [I + 1, I]-> [I + 1, I + 1]; the score of only one digit is changing at a time, which makes it easier to transfer the status. Define dp [I] to indicate the expected I + 1 score. The state transition equation is as follows:

Dp [I] = 1 × p + (1-p) * (1 + dp [I-2] + dp [I-1] + dp [I]); ans [I + 1] [I] = ans [I] [I] + dp [I], ans [I + 1] [I + 1] = ans [I + 1] [I] + dp [I];

#include <cstdio>double ans[21][21], dp[21];int main() {    double p;    int i, j;    while (~scanf("%lf", &p)) {        dp[0] = 1/p, dp[1] = 1+(1-p)/p*(dp[0]+1);        for (i = 2;i <= 19;i++) dp[i] = 1+(1-p)/p*(dp[i-2]+dp[i-1]+1);        ans[0][0] = 0,ans[1][0] = dp[0],ans[1][1] = ans[1][0]+dp[0];        for (i = 1;i <= 19;i++) {            ans[i+1][i] = ans[i][i]+dp[i];            ans[i+1][i+1] = ans[i+1][i]+dp[i];        }        printf("%.6lf\n", ans[20][19]);    }}

Acm hdu 2412 tree DP

Lzsb output Yes No, not YES NO

HDU3646 DP Solution Why WA?

Consider M> N! There is also a boundary of 0 birds and 0 monsters!
# Include <iostream>
Using namespace std;
Const int NN = 11000;
Int mn (int x, int y) {if (x <y) return x; return y ;}
Int f [NN] [102], g [NN] [102], B [NN], ksum [NN * 10];
Int main ()
{
// Freopen ("in.txt", "r", stdin );
Int pt, I, j, x, N, M, K, gg, tgg, ed, ted;
While (scanf ("% d", & N, & M, & K), N + M + K ){
For (I = 1; I <= N; I ++ ){
Scanf ("% d", & x); B [I] = x;
}
For (ksum [0] = 0, I = 1; I <= K; I ++ ){
Scanf ("% d", & x );
Ksum [I] = ksum [I-1] + x;
}
F [0] [0] = g [0] [0] = 0;

For (I = 1; I <= N; I ++ ){
For (j = 0; j <= mn (I, M); j ++ ){
Ed = ted =-1;
If (J-1> = 0 & I-1> = J-1 ){
Int st = f [I-1] [J-1];
Pt = B [I] * 2 + g [I-1] [J-1];
Ed = st;
While (ed + 1 <= K & ksum [ed + 1] <= pt ){
Ed ++;
If (ksum [ed] = pt) break;
}
If (ed = st ){
Gg = pt;
} Else gg = ksum [ed];
}
If (I-1> = j ){
Int st = f [I-1] [j];
Pt = B [I] + g [I-1] [j];
Ted = st;
While (ted + 1 <= K & ksum [ted + 1] <= pt ){
Ted ++;
If (ksum [ted] = pt) break;
}
If (ted = st ){
Tgg = pt;
} Else tgg = ksum [ted];
}
If (ted> ed | (ted = ed & tgg> gg) {gg = tgg, ed = ted ;}
F [I] [j] = ed; g [I] [j] = gg;
// Printf ("% d", f [I] [j]);
} // Printf ("\ n ");
} // Printf ("\ n ");
For (j =-1, I = 0; I <= M; I ++) if (j <f [N] [I]) j = f [N] [I];
Printf ("% d \ n", j );
}
Return 0;
}... Remaining full text>