SCU2016-02 R题概率dp

Analyse:
定义：dp[i][j][k]表示石头剩下i个，布剩下j个，剪刀剩下k个的概率.
对于当前局面可以减少一个石头，减少一个布，减少一个剪刀，和平三种情况，平了我们还会继续来一轮。
所以干脆就只考虑三种情况，他们各自发生的概率是他们的加权比例.
减少一个石头:i∗k种情况，减少一个剪刀i∗j种情况，减少一个布j∗k种情况。
减少石头的概率

i∗ki∗k+k∗j+j∗k

Get:
概率是当前局面累加到未知局面上。
期望是当前局面累加已知局面的期望天数的加权和 + 1。

/**********************jibancanyang************************** *Author*        :jibancanyang *Created Time*  : 三  7/ 6 12:44:52 2016**Problem**:**Code**:***********************[e-mail]**********************/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;typedef vector<int> vi;#define pr(x) cout << #x << ": " << x << "  " #define pl(x) cout << #x << ": " << x << endl;#define pri(a) printf("%d\n",(a))#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%lld", &(n))#define sai(n) scanf("%I64d", &(n))#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++) const int mod = int(1e9) + 7, INF = 0x3f3f3f3f;const int maxn = 1e5 + 13;double dp[102][103][104];int main(void){#ifdef LOCAL    freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    int r, s, p;    sa(r), sa(s), sa(p);    dp[r][s][p] = 1;    for (int i = r; i >= 0; i--) {        for (int j = s; j >= 0; j--) {            for (int k = p; k >= 0; k--) {                if (i * j + j * k + i * k) ;                else continue;                if (i) dp[i - 1][j][k] += dp[i][j][k] * i * k / (i * j + j * k + i * k );                 if (j) dp[i][j - 1][k] += dp[i][j][k] * i * j / (i * j + j * k + i * k );                if (k) dp[i][j][k - 1] += dp[i][j][k] * k * j / (i * j + j * k + i * k );             }        }    }    double a = 0, b = 0, c = 0;    for (int i = 1; i <= r; i++) a += dp[i][0][0];    for (int i = 1; i <= s; i++) b += dp[0][i][0];    for (int i = 1; i <= p; i++) c += dp[0][0][i];    printf("%.14f %.14f %.14f\n", a, b, c);    return 0;}