UVa 3882 And Then There Was One（stl+有技巧的模拟||数学方法+约瑟夫问题）

Let’s play a stone removing game.

Initially, n stones are arranged on a circle and numbered 1, …, n clockwise (Figure 1). You are also given two numbers k and m. From this state, remove stones one by one following the rules explained below, until only one remains. In step 1, remove stone m. In step 2, locate the k-th next stone clockwise from m and remove it. In subsequent steps, start from the slot of the stone removed in the last step, make khops clockwise on the remaining stones and remove the one you reach. In other words, skip (k − 1) remaining stones clockwise and remove the next one. Repeat this until only one stone is left and answer its number. For example, the answer for the case n= 8, k = 5, m = 3 is 1, as shown in Figure 1.

Initial state
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Step 7
Final state Figure 1: An example game

Initial state: Eight stones are arranged on a circle.

Step 1: Stone 3 is removed since m = 3.

Step 2: You start from the slot that was occupied by stone 3. You skip four stones 4, 5, 6 and 7 (since k = 5), and remove the next one, which is 8.

Step 3: You skip stones 1, 2, 4 and 5, and thus remove 6. Note that you only count stones that are still on the circle and ignore those already removed. Stone 3 is ignored in this case.

Steps 4–7: You continue until only one stone is left. Notice that in later steps when only a few stones remain, the same stone may be skipped multiple times. For example, stones 1 and 4 are skipped twice in step 7.

Final State: Finally, only one stone, 1, is on the circle. This is the final state, so the answer is 1.

Input

The input consists of multiple datasets each of which is formatted as follows.

n k m

The last dataset is followed by a line containing three zeros. Numbers in a line are separated by a single space. A dataset satisfies the following conditions.

2 ≤ n ≤ 10000, 1 ≤ k ≤ 10000, 1 ≤ m ≤ n

The number of datasets is less than 100.

Output

For each dataset, output a line containing the stone number left in the final state. No extra characters such as spaces should appear in the output.

Sample Input

8 5 3100 9999 9810000 10000 100000 0 0

Sample Output

1932019

题解：

就是类似于猴子报数问题，一共开始有n个人，标号1，2...n，从m开始，每次报数到第k个人那人就出列，然后又循环，问最后一个人的标号

以前在wustoj上用数组vis赋值遍历++的方法弱爆了，数据大一点就会超时，无限tle，我用动态数组模拟也想了一段时间了，re了几次是因为没考虑k=1的特殊情况，最后800ms过了，后来发现居然有16ms的算法。。。原来就是纯数学递推，我还是弱爆了

我的代码：

#include<algorithm>#include<iostream>#include<cstring>#include<stdio.h>#include<math.h>#include<string>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<deque>#define M (t[k].l+t[k].r)/2#define lson k*2#define rson k*2+1#define ll long longusing namespace std;vector<int>v;//存数值情况int b[10005];//存要删除数的下标int main(){    int n,k,m,i,j,len,step,t,ans,st;    while(scanf("%d%d%d",&n,&k,&m)!=EOF)    {        if(n==0&&m==0&&k==0)        {            break;        }        if(k==1)//k=1的时候要特殊处理，不然会re        {            if(m-1<1)                m=n+1;            printf("%d\n",m-1);            continue;        }        v.clear();        for(i=1;i<m;i++)//处理m前面的数        {            v.push_back(i);        }        step=0;        for(i=m+1;i<=n;i++)//处理m后面的数        {            step=(step+1)%k;//step为当前累积步数            if(!step)            {                continue;            }            v.push_back(i);        }        st=k-step-1;//st为下次循环在数组中开始的位置，稍微推一下就好了        while(v.size()>1)        {            ans=0;            for(i=st;i<v.size();i+=k)//处理要删除的数            {                b[ans]=i;//记录下标                ans++;                step=v.size()-i-1;//存当前累积步数            }            st=k-step-1;//下次开始的下标            for(i=ans-1;i>=0;i--)//要从后往前删除            {                v.erase(v.begin()+b[i]);            }            st%=v.size();//因为下标可能比数组大小大        }        printf("%d\n",v[0]);    }    return 0;}

然后有很快的数学递推的方法大佬博客讲解：http://blog.csdn.net/Baileys0530/article/details/43835589