骰子求和

骰子求和-lintcode

http://www.lintcode.com/zh-cn/problem/dices-sum/

每一次求概率都能利用上一次求得的概率

#include "map"
#define pmap map

class Solution {
public:
    /**
     * @param n an integer
     * @return a list of pair
     */
    vector> dicesSum(int n) {
        // Write your code here
        float bace = 1.0 / 6.0;
        pmap resultmap;
        vector> result;
        for(int i = 1; i <= n; ++i){
            if(i == 1){
                resultmap = {{1, bace}, {2, bace}, {3, bace}, {4, bace}, {5, bace}, {6, bace}};
            }else{
                pmap newmap;
                for(int j = 1; j <= 6; ++j){
                    for(auto &m : resultmap){
                        float newp = m.second * bace;
                        int a = m.first + j;
                        newmap[a] += newp;
                    }
                }
                resultmap = newmap;
            }
        }
        for(auto &m : resultmap){
            result.push_back({m.first, m.second});
        }
        return result;
    }
};