网址：http://acm.hdu.edu.cn/showproblem.PHP?pid=3076
题目：<h1 style="color: rgb(26,92,200); text-align: center; font-family: 'Times New Roman';">ssworld VS DDD</h1><span style="font-family: 'Times New Roman';font-size:14px; text-align: -webkit-center;"><strong><span style="font-family: Arial; font-size: 12px; color: green;">Time Limit: 4000/2000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1989Accepted Submission(s): 383 </span></strong></span><br style="font-family: 'Times New Roman';font-size:14px; text-align: -webkit-center;" /><br style="font-family: 'Times New Roman';font-size:14px; text-align: -webkit-center;" /><div class="panel_title" align="left" style="height: 38px; padding: 0px 14px; color: rgb(124,169,237); font-size: 18px; font-family: Arial; font-weight: bold; background: url(http://acm.hdu.edu.cn/images/panel-title.png) 0% 100% no-repeat transparent;">Problem Description</div><div class="panel_content" style="height: auto; margin: 0px; padding: 0px 20px; font-size: 14px; font-family: 'Times New Roman'; background: url(http://acm.hdu.edu.cn/images/panel-content.png) repeat-y;">One day,sssworld and DDD play games together,but there are some special rules in this games. They both have their own HP. Each round they dice respectively and get the points P1 and P2 (1 <= P1,P2 <= 6). Small number who,whose HP to reduce 1,the same points will remain unchanged. If one of them becomes 0 HP,he loses. As a result of technical differences between the two,each person has different probability of throwing 1,2,3,4,5,6. So we couldn’t predict who the final winner. </div><div class="panel_bottom" style="height: auto; margin: 0px; font-family: 'Times New Roman';font-size:14px; text-align: -webkit-center; background: url(http://acm.hdu.edu.cn/images/panel-bottom.png) 0% 0% no-repeat;"></div><br style="font-family: 'Times New Roman';font-size:14px; text-align: -webkit-center;" /><div class="panel_title" align="left" style="height: 38px; padding: 0px 14px; color: rgb(124,237); font-size: 18px; font-family: Arial; font-weight: bold; background: url(http://acm.hdu.edu.cn/images/panel-title.png) 0% 100% no-repeat transparent;">Input</div><div class="panel_content" style="height: auto; margin: 0px; padding: 0px 20px; font-size: 14px; font-family: 'Times New Roman'; background: url(http://acm.hdu.edu.cn/images/panel-content.png) repeat-y;">There are multiple test cases. For each case,the first line are two integer HP1,HP2 (1 <= HP1,HP2 <= 2000),said the first player sssworld’s HP and the second player DDD’s HP. The next two lines each have six floating-point numbers per line. The jth number on the ith line means the the probability of the ith player gets point j. The input data ensures that the game always has an end. </div><div class="panel_bottom" style="height: auto; margin: 0px; font-family: 'Times New Roman';font-size:14px; text-align: -webkit-center; background: url(http://acm.hdu.edu.cn/images/panel-bottom.png) 0% 0% no-repeat;"></div><br style="font-family: 'Times New Roman';font-size:14px; text-align: -webkit-center;" /><div class="panel_title" align="left" style="height: 38px; padding: 0px 14px; color: rgb(124,237); font-size: 18px; font-family: Arial; font-weight: bold; background: url(http://acm.hdu.edu.cn/images/panel-title.png) 0% 100% no-repeat transparent;">Output</div><div class="panel_content" style="height: auto; margin: 0px; padding: 0px 20px; font-size: 14px; font-family: 'Times New Roman'; background: url(http://acm.hdu.edu.cn/images/panel-content.png) repeat-y;">One float with six digits after point,indicate the probability sssworld won the game.</div><div class="panel_bottom" style="height: auto; margin: 0px; font-family: 'Times New Roman';font-size:14px; text-align: -webkit-center; background: url(http://acm.hdu.edu.cn/images/panel-bottom.png) 0% 0% no-repeat;"></div><br style="font-family: 'Times New Roman';font-size:14px; text-align: -webkit-center;" /><div class="panel_title" align="left" style="height: 38px; padding: 0px 14px; color: rgb(124,237); font-size: 18px; font-family: Arial; font-weight: bold; background: url(http://acm.hdu.edu.cn/images/panel-title.png) 0% 100% no-repeat transparent;">Sample Input</div><div class="panel_content" style="height: auto; margin: 0px; padding: 0px 20px; font-size: 14px; font-family: 'Times New Roman'; background: url(http://acm.hdu.edu.cn/images/panel-content.png) repeat-y;"><pre style="word-wrap: break-word; white-space: pre-wrap; margin-top: 0px; margin-bottom: 0px;"><div style="font-family: 'Courier New',Courier,monospace;">5 5 1.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 1.000 5 5 0.000 0.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 0.000 0.000 0.000</div>
Sample Output
0.000000 1.000000
Source
题意A、B两人扔色子，分别给出他们初始的分数H1和H2，再给出它们扔 1- 6的概率，扔出分数低的人会失去一分，平局则继续扔，最先变为0的人就输，比赛终止。
当一个人扔出的分数不另一个人小时，他就会输，所以利用两层循环找出A，B各自获胜的概率wa，wb，见代码。若果出现平局，他们会继续扔，那么A,B各自获胜的概率依然是
wa，wb，一直循环下去；所以每一句A、B获胜的概率就是s=wa+wb; wa=wa/s;(A的） wb=wb/s;（B的）；
然后用dp[i][j]表示A分数为i，b分数为j是A获胜的概率，因为一局只有A胜或B胜2总情况，所以dp[i][j]=dp[i-1][j]*wb+dp[i][j-1]*wa;初始化B的分数为0时，dp[i][0]=1;
又dp[0][j]=0;
最后题目内存限制采用滚动数组；最坑的是目数据H1，H2是反的，一直wrong到上网看题解，要反过来才行。
#define _CRT_SECURE_NO_DEPRECATE #include <iostream> #include <cstdio> #include<cstdlib> #include <cstring> #include <algorithm> #include <set> #include <string> #include <cctype> #include <map> #include <vector> #include<stack> #include<cmath> #include<queue> #define pb push_back #define mp make_pair #define mem(a) memset(a,sizeof(a)) #define fore(i,s,t) for(int i=s;i<t;i++) #define LL __int64 using namespace std; const int maxn = 2100,inf = ~(1 << 31),mod = 258280327; const double eps = 1e-7,pi = acos(-1.0); double dp[maxn],adp[maxn],pa[7],pb[7],wa,wb; int main(void) { int n,m; while(scanf("%d %d",&n,&m)!=EOF) { mem(dp); mem(adp); wa=wb=0; for(int i=1; i<=6; i++)scanf("%lf",&pa[i]); for(int i=1; i<=6; i++)scanf("%lf",&pb[i]); for(int i=2; i<=6; i++) for(int j=1; j<i; j++) { wa+=pa[i]*pb[j]; wb+=pb[i]*pa[j]; } double s=wa+wb; wa/=s; wb/=s; for(int i=1; i<=m; i++) { for(int j=1; j<=n; j++) dp[j]=adp[j]*wb+(j==1?1:dp[j-1])*wa; for(int j=1; j<=n; j++) adp[j]=dp[j]; } printf("%.6lf\n",dp[n]); } return 0; } /* 0.200 0.300 0.200 0.000 0.000 0.300 0.000 0.000 0.500 0.000 0.000 0.500 2 1 0.200 0.300 0.200 0.000 0.000 0.300 0.000 0.000 0.500 0.000 0.000 0.500 1 2 0.000 0.000 0.500 0.000 0.000 0.500 0.200 0.300 0.200 0.000 0.000 0.300 1 1000 0.000 0.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 0.000 0.000 0.000 */原文链接：https://www.f2er.com/javaschema/284507.html
