题目意思是，AB两个人掷硬币，每次一个人掷两次，然后对应图标里面得分，要你输出前20回合 A赢，B赢，或是平均的概率
dp还是不怎么会，参考别人代码才敲出来的
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个回合，A得j分，B得k分的概率，j或k最小为-20，所以+20 20 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 "); 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%% ",i,a_win*100,b_win*100,tie*100); 44 } 45 } 46 int main() 47 { 48 solve(); 49 }