题意：
有2^n支队伍进行比赛，每行给出这支队伍打败各支队伍的几率，求获胜几率最大的队伍
1 #include<iostream> 2 #include<string> 3 #include<cstdio> 4 #include<vector> 5 #include<queue> 6 #include<stack> 7 #include<algorithm> 8 #include<cstring> 9 #include<stdlib.h> 10 #include<cmath> 11 using namespace std; 12 #define pb push_back 13 double p[140][140]; 14 double dp[10][130]; 15 int need(int x,int ps){ 16 if(x%(1<<(ps))>=1&&x%(1<<(ps))<=(1<<(ps-1))) 17 return x/(1<<ps)*(1<<ps)+(1<<(ps-1))+1; 18 if(x%(1<<(ps))==0) 19 return (x/(1<<ps)-1)*(1<<ps)+1; 20 return x/(1<<ps)*(1<<ps)+1; 21 } 22 int main(){ 23 int n; 24 while(cin>>n&&n!=-1){ 25 for(int i=1;i<=(1<<n);i++) 26 for(int j=1;j<=(1<<n);j++) 27 scanf("%lf",&p[i][j]); 28 memset(dp,0,sizeof(dp)); 29 for(int i=1;i<=(1<<n);i++) 30 dp[0][i]=1; 31 for(int i=1;i<=n;i++){ 32 for(int j=1;j<=(1<<n);j++){ 33 int kk=need(j,i); 34 for(int k=kk;k<kk+(1<<(i-1));k++) 35 dp[i][j]+=dp[i-1][j]*dp[i-1][k]*p[j][k]; 36 } 37 } 38 int pos; 39 double mm=0; 40 for(int i=1;i<=(1<<n);i++) 41 if(dp[n][i]>mm){ 42 mm=dp[n][i]; 43 pos=i; 44 } 45 cout<<pos<<endl; 46 } 47 }