Ultraviolet A 1541-to bet or not to bet (memory + probability)

Link to the question: Ultraviolet A 1541-to bet or not to bet

A game is played on a board. Given the length of the board m, it is not counted as the start and end, and the number of steps T. From the starting point, one coin can be lost in each round, and two steps can be moved on the front. One step is moved on the front. The middle lattice has a write operation, including moving a certain number of steps to stop an operation. The probability of reaching the end point in step T.

Solution: DP [I] [J] indicates the probability that step J is used to get to the I lattice, and then the memory is searched because the State is repeated and recursive termination conditions can be determined.

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int maxn = 100;bool lose[maxn], vis[maxn][maxn];double p[maxn][maxn];int N, T, inst[maxn];void init () {    char str[maxn];    memset(inst, 0, sizeof(inst));    memset(vis, false, sizeof(vis));    memset(lose, false, sizeof(lose));    scanf("%d%d", &N, &T);    for (int i = 1; i <= N; i++) {        scanf("%s", str);        if (str[0] == ‘L‘)            lose[i] = true;        else            sscanf(str, "%d", &inst[i]);    }}double dfs (int d, int k) {    if (vis[d][k])        return p[d][k];    vis[d][k] = true;    if (d == N + 1)        return p[d][k] = 1;    if (k <= 0)        return p[d][k] = 0;    double& ret = p[d][k];    ret = 0;    int next = d + 1;    if (lose[next])        ret += 0.5 * dfs(next, k-2);    else        ret += 0.5 * dfs(next + inst[next], k - 1);    next = min(d + 2, N + 1);    if (lose[next])        ret += 0.5 * dfs(next, k-2);    else        ret += 0.5 * dfs(next + inst[next], k - 1);    return ret;}int main () {    int cas;    scanf("%d", &cas);    while (cas--) {        init();        double ans = dfs(0, T);        if (fabs(ans - 0.5) < 1e-9)            printf("Push. 0.5000\n");        else if (ans > 0.5)            printf("Bet for. %.4lf\n", ans);        else            printf("Bet against. %.4lf\n", ans);    }    return 0;}