leetcode-剑指60-OK

// language c
// 剑指60
// https://leetcode-cn.com/problems/nge-tou-zi-de-dian-shu-lcof/


/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
double* dicesProbability(int n, int* returnSize){
	double ans[6*n+1];	// 用来暂时存储答案
	ans[1] = 1.0/6.0;
	ans[2] = 1.0/6.0;
	ans[3] = 1.0/6.0;
	ans[4] = 1.0/6.0;
	ans[5] = 1.0/6.0;
	ans[6] = 1.0/6.0;      //第一波


	// 检查是b否在a的圈内,在的话返回true
	bool check(int a,int b){
		if((b >= a) && (b <=6*a))
			return true;
		return false;
	}


	for(int i =2; i<=n; i++){	// 每次循环是算i个骰子的答案，会占有i~6i的位置，i从2算到n

		double bns[5*i+1];	// 用来暂时存储, 0号存的值代表i的概率
		for(int j=0; j<5*i+1; j++)		// 初始化
			bns[j] = 0.0;

		for(int j = i; j<=6*i; j++){	// 算某个和的概率
			for( int k = 1; k<=6; k++){ 
				if(check(i-1,j-k))
					bns[j-i] += ans[j-k]/6.0;
			}
		}

		for(int j=0; j<5*i+1; j++)		// 填回去
			ans[i+j] = bns[j];
	}


	// 填回去了，准备返回答案，这步不会错的
	returnSize[0] = 5*n+1;
	double* AAA = (double *)malloc(sizeof(double) * (5*n+1));
	for(int i = 0; i<5*n+1; i++)
		AAA[i] = ans[i+n];

	return AAA;
}