Codeforces 111B【看看自己和别人在代码能力上的差距！】

我的：

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

int x[100001],y[100001],d[100001];
int i =1;

int main()
{
    memset(d,-1,sizeof(d));
    int n;
    cin>>n;
    while(n--)
    {
        int ans = 0,m;
        cin>>x[i]>>y[i];
        for(int j = 1; j*j <= x[i]; j++)
        {
            if(x[i]%j == 0)
            {
                ans++;
                m = j;
            }
        }
        if(m * m == x[i])
            ans = ans*2-1;
        else
            ans = ans*2;
        if(y[i])
        {
            for(int j = 1; j*j <= x[i]; j++)
            {
                if(x[i] % j == 0 && i - d[j] <= y[i])
                    ans--;
                if(x[i] % (x[i]/j) == 0 && i - d[x[i]/j] <= y[i] && j != x[i]/j)
                    ans--;
            }
            cout<<ans<<endl;
        }
        else if(y[i] == 0)
            cout<<ans<<endl;
        for(int j = 1; j*j <= x[i]; j++)
        {
            if(x[i]%j == 0)
            {
                d[j] = i;
                d[x[i]/j] = i;
            }
        }
        i++;
    }
}

别人的：

/*#include<iostream>
#include<stack>
#include<queue>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<cmath>
#define ll long long
#define oo 1000000007
#define MAXN 100005
using namespace std;
int last[MAXN];
int main()
{
      int T,t;
      scanf("%d",&T);
      memset(last,-1,sizeof(last));
      for (t=1;t<=T;t++)
      {
             int x,y,ans,i;
             ans=0;
             scanf("%d%d",&x,&y);
             for (i=1;i*i<=x;i++)
               if (x%i==0)
               {
                     if (t-last[i]>y) ans++;
                     if (x-i*i && t-last[x/i]>y) ans++;
                     last[i]=last[x/i]=t;
               }
             printf("%d
",ans);
      }
      return 0;
}
*/
#include<cstdio>
#include<cstring>
#define N 100100

int d[N],n,x,t;

int main()
{
    memset(d,-1,sizeof(d));
    scanf("%d",&t);
    for(int idx=1;idx<=t;idx++) {
        scanf("%d%d",&x,&n);
        int ans=0;
        for(int i=1;i*i<=x;i++) {
            if((x%i)==0) {
                int c1=i;
                int c2=x/i;
                //一次可以找到两个约数，i和x/i；
                if(d[c1]+n<idx) ans++;
                if(c1!=c2 && d[c2]+n<idx) ans++;
                d[c1]=d[c2]=idx;
            }
        }
        printf("%d
",ans);
    }
    return 0;
}