动态规划求一定数量骰子和的概率

2018华为实习机试

题目描述：略

动态规划：

f(n,s)=f(n-1,s-1)+f(n-1,s-2)+f(n-1,s-3)+.....+f(n-1,s-6)      至于f(n-1,s-k)是否存在，由Record的getinfo()判断，不存在就设为０

#include <iostream>
#include<vector>
#include<iomanip>
#include<cmath>

using namespace std;
class Record{
public:
    vector<vector<int>> data;
    Record(){}
    Record(int s)
    {
        data.resize(s);
        for(int i=0;i<s;++i){
            data[i].resize(5*i+6);
        }
    }
    bool addinfo(int dicenum,int sum,int cnt)
    {
        if(sum>=dicenum && sum<=dicenum*6){
            data[dicenum-1][sum-dicenum]=cnt;
            return true;
        }
        else{
            return false;
        }
    }
    int getinfo(int dicenum,int sum)
    {
        if(sum>=dicenum && sum<=dicenum*6){
            return data[dicenum-1][sum-dicenum];
        }
        else{
            return 0;
        }
    }
};


int main()
{
    int dicenum;
    int total;
    while(cin >> dicenum){
        int smin,smax;
        Record rec(dicenum);
        total = (int)(0.5+pow(6,dicenum));
        for(int i=1;i<=dicenum;++i){
            smin = i;
            smax = i*6;
            if(i==1){
                for(int j=smin;j<=smax;++j){
                    rec.addinfo(1,j,1);
                }
            }
            else{
                for(int j=smin;j<=smax;++j){
                    int tempcnt = 0;
                    for(int k=1;k<=6;++k){
                        tempcnt += rec.getinfo(i-1,j-k);
                    }
                    rec.addinfo(i,j,tempcnt);
                }
            }
        }
        /*test rec data
        for(int i=1;i<=dicenum;++i){
            int temp = 0;
            cout << i << endl;
            for(int j=i;j<=i*6;++j){
                if((temp=rec.getinfo(i,j))!=0){
                    cout << temp << " ";
                }
            }
            cout << endl;
        }
        */

        /*
        output 1: 当数字很大时，结果不对称
        double prob = 0;
        cout << "[";
        for(int i=smin;i<=smax;++i){
            prob = (double)(rec.getinfo(dicenum,i)) / total;
            cout << "[" << i << ", ";
            cout.flags(ios::fixed);
            cout.precision(5);
            cout << prob << "]";
            if(i != smax){
                cout << ", ";
            }
        }
        cout << "]" << endl;
        */

        //output 2:使输出结果对称，且减少运算量
        vector<double> prob;
        prob.resize((smax+smin)/2);
        cout << "[";
        for(int i=smin;i<=(smin+smax)/2;++i){
            prob[i-smin] = (double)(rec.getinfo(dicenum,i)) / total;
            cout << "[" << i << ", ";
            cout.flags(ios::fixed);
            cout.precision(5);
            cout << prob[i-smin] << "]" << ", ";
        }
        int tempsum = (smin+smax)/2+1;
        for(int i=prob.size()-1;i>=0;--i){
            if(i==prob.size()-1){
                if((smin+smax)%2!=0){
                    cout << "[" << tempsum << ", ";
                    cout.flags(ios::fixed);
                    cout.precision(5);
                    cout << prob[i] << "]" << ", ";
                    tempsum++;
                }
            }
            else{
                cout << "[" << tempsum << ", ";
                cout.flags(ios::fixed);
                cout.precision(5);
                cout << prob[i] << "]";
                tempsum++;
                if(i!=0){
                    cout << "], ";
                }
            }
        }
        cout << "]" << endl;
    }
    return 0;
}

 程序缺陷：对于多个测例，没有将之前Record类的信息保留节省计算