Codeforces 167b wizards and huge prize probability dp (Water

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Question:

Given n opponents, at least one of them will be defeated. Now there is a pocket CAPACITY OF K

The N numbers below indicate the probability of defeating this person.

The following n numbers (if-1 indicates that a gold coin can be obtained if this person is defeated, and if it is greater than 0, the pocket capacity can be increased to this number)

Q:

Beat at least one of them, and the total pocket capacity obtained> = the probability of the number of gold coins. (That is, gold coins cannot be stored at any time)

Ideas:

Probability DP

Note that it is possible that the pocket capacity is negative, but it turns into a positive number at the end, so we need to increase the pocket capacity by N,

# Include <iostream> # include <map> # include <string> # include <cstring> # include <cstdio> template <class T> inline bool RD (T & RET) {char C; int SGN; If (C = getchar (), c = EOF) return 0; while (C! = '-' & (C <'0' | C> '9') c = getchar (); SGN = (C = '-')? -1:1; ret = (C = '-')? 0 :( C-'0'); While (C = getchar (), c> = '0' & C <= '9 ') ret = RET * 10 + (c-'0'); RET * = SGN; return 1;} template <class T> inline void Pt (t x) {If (x <0) {putchar ('-'); X =-X;} If (x> 9) Pt (X/10 ); putchar (X % 10 + '0');} using namespace STD; const int n = 201; const double EPS = 1e-10; int n, l, D; double P [N], DP [N] [N] [N * 2 + 10]; // (I, j, k) indicates that the first one hits J, probability int B [N]; void input () {for (INT I = 1, J; I <= N; I ++) {RD (j); P [I] = (double) J/100.0;} For (INT I = 1; I <= N; I ++) rd (B [I]);} int main () {While (CIN> N> L> d) {input (); memset (DP, 0, sizeof DP); DP [0] [0] [d + N] = 1.0; For (INT I = 1; I <= N; I ++) for (Int J = 0; j <I; j ++) for (int K = 0; k <= 2 * n; k ++) {If (B [I] =-1) {If (k> 0) DP [I] [J + 1] [k-1] + = DP [I-1] [J] [k] * P [I];} else {If (K + B [I]> N * 2) DP [I] [J + 1] [N * 2] + = DP [I-1] [J] [k] * P [I]; elsedp [I] [J + 1] [K + B [I] + = DP [I-1] [J] [k] * P [I];} DP [I] [J] [k] + = DP [I-1] [J] [k] * (1-p [I]);} double ans = 0; for (INT I = L; I <= N; I ++) for (Int J = N; j <= 2 * n; j ++) ans + = DP [N] [I] [J]; printf ("%. 10f \ n ", ANS);} return 0 ;}

Codeforces 167b wizards and huge prize probability dp (Water