A - Chinese Girls' Amusement ZOJ - 2313(大数)

You must have heard that the Chinese culture is quite different from that of Europe or Russia. So some Chinese habits seem quite unusual or even weird to us.

So it is known that there is one popular game of Chinese girls. N girls stand forming a circle and throw a ball to each other. First girl holding a ball throws it to the K-th girl on her left (1 <= K <= N/2). That girl catches the ball and in turn throws it to the K-th girl on her left, and so on. So the ball is passed from one girl to another until it comes back to the first girl. If for example N = 7 and K = 3, the girls receive the ball in the following order: 1, 4, 7, 3, 6, 2, 5, 1.

To make the game even more interesting the girls want to choose K as large as possible, but they want one condition to hold: each girl must own the ball during the game.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Input
Input file contains one integer number N (3 <= N <= 10^2000) - the number of Chinese girls taking part in the game.

Output
Output the only number - K that they should choose.

Sample Input
2

7

6

Sample Output
3

1

//大数加和大数除再加上找规律
//总体上分为两种情况:
//(1) 如果n是奇数的话,K=(n-1)/2
//(2) 如果n是偶数又分两种情况:(2.1) 如果n/2是偶数的话 K=n/2-1 (2.2) 如果n/2是奇数的话 K=n/2-2

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<cstdio>
#include<sstream>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include <ctype.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;

typedef long long ll;
const int maxn=110000;
const int INF=0x3f3f3f3f;

void jminus(int a[],int len,int b)
{
    a[len-1]-=b;
    for(int i=len-1; i>0; i--)//借位
    {
        if(a[i]<0)
        {
            a[i]+=10;
            a[i-1]-=1;
        }
    }
}
void divide(int a[],int len )//大数除
{
    for(int i=0; i<len-1; i++)
    {
        if(a[i]%2==0)  a[i]=a[i]/2;
        else
        {
            a[i]=a[i]/2;
            a[i+1]+=10;
        }
    }
    a[len-1]/=2;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        char str[2010];
        cin>>str;
        int len=strlen(str);
        int a[len];

        for(int i=0; i<len; i++)
            a[i]=str[i]-'0';

        if(a[len-1]%2==1)
        {
            divide(a,len);
            int start=0;
            while(a[start]==0)  start++;

            for(int i=start; i<len; i++)
                cout<<a[i];
            cout<<endl;
        }
        else
        {
            if(len==1)
            {
                int temp=a[0];
                if(temp%4==2)//n是偶数,n/2是奇数的情况
                {
                    divide(a,len);
                    jminus(a,len,2);
                }
                else//n是偶数,n/2是偶数
                {
                    divide(a,len);
                    jminus(a,len,1);
                }
            }
            else
            {
                int temp=a[len-1]+10*a[len-2];
                if(temp%4==2)
                {
                    divide(a,len);
                    jminus(a,len,2);
                }
                else
                {
                    divide(a,len);
                    jminus(a,len,1);
                }
            }
            int start=0;
            while(a[start]==0)  start++;

            for(int i=start; i<len; i++)
                cout<<a[i];
            cout<<endl;
        }
        if(t>0)
            cout<<endl;
    }
    return 0;
}