n個骰子，求所有可能和出現的概率&&約瑟夫環

有兩種解法：

一、遞歸求出所有的可能性，然後計算概率。

二、遞歸過程需要重複計算，所以可以用數組來存儲結果提升速度。

下面代碼有兩種方式並可以看出時間的差異。

#include<iostream>
#include<string.h>
#include<cmath>
#include<time.h>
using namespace std;

void basegetdicechance(int *totalsum,int current,int sum,int k){
	if(current == 0){
		++totalsum[sum-k];
		return;
	}
	for(int i =1;i<=6;++i){
		basegetdicechance(totalsum,current-1,sum+i,k);
	}
}

int main(){
	int k;
	clock_t start,end;
	printf("獲得骰子的個數:\n");
	scanf("%d",&k);
	int *totalsum = (int *)malloc(sizeof(int)*(5*k+1));
	for(int i =0;i<=5*k;++i)
		totalsum[i] = 0;
	start = clock();
	double sumchance = pow(6.0,(double)k);
	basegetdicechance(totalsum,k,0,k);
	for(int i =0;i<=5*k;++i){
		printf("基本方法：和爲%d的概率出現爲百分之%f\n",i+k,(double)totalsum[i]*100/sumchance);
	}
	end = clock();
	printf("基本方法所需時間爲%f\n",(double)(end-start)/CLOCKS_PER_SEC);
	free(totalsum);
	//快速的方法
	int *totaltwosum[2];
	totaltwosum[0] = (int *)malloc(sizeof(int)*(6*k+1));
	totaltwosum[1] = (int *)malloc(sizeof(int)*(6*k+1));
	for(int i =0;i<=6*k;++i){
		totaltwosum[0][i] =0;
		totaltwosum[1][i] =0;
	}
	start = clock();
	int flag = 0;
	for(int i =1;i<=6;++i)
		totaltwosum[flag][i] = 1;
	for(int j =2;j<=k;j++){
		for(int i = 0;i<j;++i)
			totaltwosum[1-flag][i] =0;
		for(int sum =j;sum <=6*j;sum++){
			totaltwosum[1-flag][sum] = 0;
			for(int i = 1;i<=sum&&i<=6;++i)
				totaltwosum[1-flag][sum] += totaltwosum[flag][sum-i];
		}
		flag = 1-flag;
	}
	for(int i =k;i<=6*k;++i){
		printf("快速方法：和爲%d的概率出現爲百分之%f\n",i,(double)totaltwosum[flag][i]*100/sumchance);
	}
	end = clock();
	printf("快速方法所需時間爲%f\n",(double)(end-start)/CLOCKS_PER_SEC);
	system("PAUSE");
	return 0;
}

約瑟夫環，比較簡單的問題了，用鏈表來解。

#include<iostream>
#include<list>
using namespace std;

int getlastone_list(unsigned int n,unsigned int m){
	list<int> jopher;
	for(int i =0;i<n;++i)
		jopher.push_back(i);
	list<int>::iterator current = jopher.begin();
	while(jopher.size()>1){
		for(int i =1;i<m;++i){
			++current;
			if(current == jopher.end())
				current = jopher.begin();
		}
		printf(" %d",*current);
		current = jopher.erase(current);
		if(current == jopher.end())
			current = jopher.begin();
		/*這裏涉及erase之後迭代器失效，而erase方法會返回下個迭代器*/
	}
	printf("\n");
	return *current;
}

int main(){
	printf("輸人n和m:\n");
	unsigned int n,m;
	scanf("%d%d",&n,&m);
	int result = getlastone_list(n,m);
	printf("最後出來的是%d",result);
	system("PAUSE");
	return 0;
}

至於另外一種O(n)的更復雜方法，數學證明比較難。略。