POJ 1217 FOUR QUARTERS

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<set>
#include<vector>
#include<map>
#include<algorithm>
#include<cmath>
#include<stdlib.h>
using namespace std;
double way[9]= {0.0625,0.125,0.0625,0.125,0.25,0.125,0.0625,0.125,0.0625};
int a[9]= {1,1,2,0,0,1,-1,0,0};
int b[9]= {0,-1,-1,1,0,0,2,0,-1};
double dp[22][66][66];
void solve()
{
    memset(dp,0,sizeof(dp));
    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
    for(int i=1; i<=20; i++)
        for(int j=60; j>=0; j--){
            int score1=j-20;
            for(int k=60; k>=0; k--){
                int score2=k-20;
                if(dp[i-1][j][k]>0){
                    for(int s=0; s<9; s++)
                        dp[i][score1+20+a[s]][score2+20+b[s] ]+=dp[i-1][j][k]*way[s];
                }

            }
        }
    printf("Round   A wins    B wins    Tie
");
    for(int i=1; i<=20; i++)
    {
        double a_win = 0 , b_win = 0 , tie = 0 ;
        for(int j=0; j<=60; j++){
            for(int k=0; k<=60; k++){
                if(j>k) a_win+=dp[i][j][k] ;
                else if(j<k)    b_win += dp[i][j][k] ;
                else tie += dp[i][j][k] ;
            }
        }
        printf("%5d%10.4f%%%9.4f%%%9.4f%%
",i,a_win*100,b_win*100,tie*100);
    }
}
int main()
{
    solve();
}