标签:概率dp namespace cstring 相等 idt ++ map inpu printf
问从0出发到达n或超过n摇色子的次数的期望。
#include <cstdio> #include <iostream> #include <cstring> #include <queue> #include <algorithm> #include <map> #include <cmath> 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; typedef long long ll; const int N = 100010; int n, m; double dp[N]; int to[N]; double solve(){ for(int i = 0; i < 6; i++) dp[i+n] = 0; for(int i = n-1; i >= 0; i--) { if(to[i]!=-1) dp[i] = dp[to[i]]; else { dp[i] = 1.0; for(int j = 1; j <= 6; j++) dp[i] += dp[i+j] / 6.0; } } return dp[0]; } void input(){ memset(to, -1, sizeof to); while(m--){ int x, y;rd(x); rd(y); to[x] = y; } for(int i = n-1; i >= 0; i--){ if(to[i] == -1)continue; if(to[to[i]] != -1) to[i] = to[to[i]]; } } int main() { while(cin>>n>>m, n+m){ input(); printf("%.4f\n", solve()); } return 0; } /* */
标签:概率dp namespace cstring 相等 idt ++ map inpu printf
原文地址:http://www.cnblogs.com/ljbguanli/p/7300685.html