标签:style blog color io os for 数据 div sp
直接看代码就OK。思路比较简单。就是注意概率要在转移过程中算出来。不能算成成立的方案书除以总方案数(POJ的这道题可以这么干。数据很水么。另外POJ要用%.5f,%.5lf 会WA。)
#include <map> #include <set> #include <list> #include <cmath> #include <ctime> #include <deque> #include <stack> #include <queue> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <climits> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define LL long long #define PI 3.1415926535897932626 using namespace std; int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);} double dp[105][15]; int N,K; void slove() { if (K <= 1) {puts("100.00000");return;} for (int i = 0 ;i < 105; i++) for (int j = 0;j <15; j++) dp[i][j]=0.0; for (int i = 0 ;i <= K; i++) dp[1][i]=100.0/(double)(K+1); for (int i = 2; i <= N; i++) { dp[i][0] = 1.0/(double)(K+1) * ( dp[i-1][0] + dp[i-1][1]); for (int j = 1; j <= K ; j++) { if (j == K) dp[i][j] = 1.0/(double)(K+1) * (dp[i-1][K] + dp[i-1][K-1]); else dp[i][j] = 1.0/(double)(K+1) * (dp[i-1][j-1] + dp[i-1][j] + dp[i-1][j+1]); } } double ans=0.0; for (int i = 0 ;i <= K; i ++) ans+=dp[N][i]; printf("%.5lf\n",ans); } int main() { while (scanf("%d%d",&K,&N)!=EOF) slove(); return 0; }
UVA 10081 Tight numbers(POJ 2537)
标签:style blog color io os for 数据 div sp
原文地址:http://www.cnblogs.com/Commence/p/3989077.html