POJ 1217 FOUR QUARTERS

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題目意思是，AB兩個人擲硬幣，每次一個人擲兩次，然後對應表徵圖裡面得分，要你輸出前20回合 A贏，B贏，或是平均的機率

dp還是不怎麼會，參考別人代碼才敲出來的

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<string> 5 #include<set> 6 #include<vector> 7 #include<map> 8 #include<algorithm> 9 #include<cmath>10 #include<stdlib.h>11 using namespace std;12 double way[9]= {0.0625,0.125,0.0625,0.125,0.25,0.125,0.0625,0.125,0.0625};13 int a[9]= {1,1,2,0,0,1,-1,0,0};14 int b[9]= {0,-1,-1,1,0,0,2,0,-1};15 double dp[22][66][66];16 void solve()17 {18     memset(dp,0,sizeof(dp));19     dp[0][20][20]=1;      //dp[i][j][k],表示第i個回合，A得j分，B得k分的機率，j或k最小為-20，所以+2020     for(int i=1; i<=20; i++)21         for(int j=60; j>=0; j--){22             int score1=j-20;23             for(int k=60; k>=0; k--){24                 int score2=k-20;25                 if(dp[i-1][j][k]>0){26                     for(int s=0; s<9; s++)27                         dp[i][score1+20+a[s]][score2+20+b[s] ]+=dp[i-1][j][k]*way[s];28                 }29 30             }31         }32     printf("Round   A wins    B wins    Tie\n");33     for(int i=1; i<=20; i++)34     {35         double a_win = 0 , b_win = 0 , tie = 0 ;36         for(int j=0; j<=60; j++){37             for(int k=0; k<=60; k++){38                 if(j>k) a_win+=dp[i][j][k] ;39                 else if(j<k)    b_win += dp[i][j][k] ;40                 else tie += dp[i][j][k] ;41             }42         }43         printf("%5d%10.4f%%%9.4f%%%9.4f%%\n",i,a_win*100,b_win*100,tie*100);44     }45 }46 int main()47 {48     solve();49 }