【NOIP 2016 提高组】玩具谜题

Description

Solution

直接用加法减法+模即可。

复杂度： O(n)

这么水的题我写来干嘛

Code

#include
#include
#include
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int N=100500;
int n,m,ans;
int a[N];
int nm[N][15];
int main()
{
    freopen("toy.in","r",stdin);
    freopen("toy.out","w",stdout);
    int q,w;char ch;
    scanf("%d%d",&n,&m);
    fo(i,0,n-1)
    {
        scanf("%d",&q);
        a[i]=1;if(q)a[i]=-1;
        ch=' ';
        for(ch=' ';ch<'a'||ch>'z';ch=getchar());
        while(ch<='z'&&ch>='a')nm[i][++nm[i][0]]=ch,ch=getchar();
    }
    ans=0;
    fo(i,1,m)
    {
        scanf("%d%d",&q,&w);
        if(q)q=1;else q=-1;
        ans=(ans+q*a[ans]*w+n)%n;
    }
    fo(i,1,nm[ans][0])printf("%c",nm[ans][i]);
    return 0;
}