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codeforces 641 C Little Artem and Dance

时间:2016-05-06 21:52:32      阅读:179      评论:0      收藏:0      [点我收藏+]

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分析:算1和2的起点在哪里,然后往后累加就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>

using namespace std;

int ans[1000005];

int main()
{
    //freopen("1.in","r", stdin);
    //freopen("1.txt","w",stdout);
    int n, m;
    while(~scanf("%d %d", &n, &m))
    {
        int x, y;
        int sum1=1, sum2=2;
        for(int i=1; i<=m; i++)
        {
            scanf("%d", &x);
            if(x==1)
            {
                scanf("%d", &y);
                sum1+=y;
                sum2+=y;
                sum1%=n;sum1+=n;sum1%=n;
                sum2%=n;sum2+=n;sum2%=n;
            }
            else
            {
                if(sum2%2)
                {
                    sum1--;
                    sum2++;
                }
                else
                {
                    sum1++;
                    sum2--;
                }
                sum1%=n;sum1+=n;sum1%=n;
                sum2%=n;sum2+=n;sum2%=n;
            }
        }
        sum1%=n;sum1+=n;sum1%=n;
        sum2%=n;sum2+=n;sum2%=n;
        for(int i=1; i<=n; )
        {
            if(sum1==0)
                sum1 = n;
            if(sum2==0)
                sum2 = n;
            ans[sum1] = i++;
            ans[sum2] = i++;
            sum1=(sum1+2+n)%n;
            sum2=(sum2+2+n)%n;
        }
        for(int i=1; i<=n; i++)
            printf("%d%c", ans[i],i==n?\n: );
    }

    return 0;
}

 

codeforces 641 C Little Artem and Dance

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原文地址:http://www.cnblogs.com/mengzhong/p/5467044.html

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