gcj 2008 apac problem c

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#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define MAX_M 25using namespace std;typedef long long ll;int M,X;int T,temp;double P;double dp[2][(1<<MAX_M)+1];void solve(){    int n=1<<M;    double *prv =dp[0], *nxt=dp[1];    memset(prv,0,sizeof(double)*(n+1));    prv[n]=1.0;    for(int r=0;r<M;r++)    {        for(int i=0;i<=n;i++)        {            int jub=min(i,n-i);            double t=0.0;            for(int j=0;j<=jub;j++)            {                t=max(t,P*prv[i+j]+(1-P)*prv[i-j]);            }            nxt[i]=t;        }        swap(prv,nxt);    }    int i=(ll) X*n/1000000;    printf("Case #%d: %.6lf\n",temp-T,prv[i]);}int main(){    freopen("/Users/xxxxxx/Desktop/input.txt","r",stdin);    freopen("/Users/xxxxxx/Desktop/output.txt","w",stdout);    cin>>T;    temp=T;    while(T--)    {        scanf("%d%lf%d",&M,&P,&X);        solve();    }    return 0;}

看的挑战才写出来注意在M>=2的时候你的策略有很多，所以钱可以从0-2*当前格子但要保证不超过n 所以用jub 来限制循环的范围

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