> #include <iostream>
> using namespace std;
> 
> long long a[40000];
> 
> void solve(int t, int j)
> {
>     int tem = t-a[j], tempo;
>     if (tem < 10)
>     {
>         printf("%d\n",tem);
>         return;
>     }
>     else if (tem >= 10 && tem < 190)
>     {
>         tempo = tem-9;
>         tempo = (tempo+1)/2+9;
>         if (tem%2)
>         {
>             printf("%d\n",tempo%10);
>             return;
>         }
>         else
>         {
>             printf("%d\n",tempo/10);
>             return;
>         }
>     }
>     else if (tem >= 190 && tem < 2890)
>     {
>         tempo = tem-189;
>         tempo = (tempo+2)/3+99;
>         switch(tem%3)
>         {
>             case 0:
>                 printf("%d\n",tempo%10);
>                 return;
>             case 1:
>                 printf("%d\n",tempo/100);
>                 return;
>             case 2:
>                 printf("%d\n",tempo/10%10);
>                 return;
>             default: return;
>         }
>     }
>     else if (tem >= 2890 && tem < 38890)
>     {
>         tempo = tem-2889;
>         tempo = (tempo+3)/4+999;
>         switch(tem%4)
>         {
>             case 2:
>                 printf("%d\n",tempo/1000);
>                 return;
>             case 3:
>                 printf("%d\n",tempo/100%10);
>                 return;
>             case 0:
>                 printf("%d\n",tempo/10%10);
>                 return;
>             case 1:
>                 printf("%d\n",tempo%10);
>                 return;
>             default: return;
>         }
>     }
>     else
>     {
>         tempo = tem - 38889;
>         tempo = (tempo+4)/5 + 9999;
>         switch(tem%5)
>         {
>             case 0:
>                 printf("%d\n",tempo/10000);
>                 return;
>             case 1:
>                 printf("%d\n",tempo/1000%10);
>                 return;
>             case 2:
>                 printf("%d\n",tempo/100%10);
>                 return;
>             case 3:
>                 printf("%d\n",tempo/10%10);
>                 return;
>             case 4:
>                 printf("%d\n",tempo%10);
>                 return;
>         }
>     }
> }
> int main()
> {
>     int N, t;
>     unsigned int i, j;
>     scanf("%d",&N);
>     a[0] = 0;
>     for (i = 1; i < 10; i++)
>     {
>         a[i] = a[i-1]+i;
>     }
>     for (i = 10; i < 100; i++)
>     {
>         a[i] = a[i-1]+2*i-9;
>     }
>     for (i = 100; i < 1000; i++)
>     {
>         a[i] = a[i-1]+3*(i-99)+189;
>     }
>     for (i = 1000; i < 10000; i++)
>     {
>         a[i] = a[i-1]+4*(i-999)+2889;
>     }
>     for (i = 10000; a[i-1] <= 2147483647; i++)
>     {
>         a[i] = a[i-1]+4*(i-9999)+38889;
>     }
>     for (int i = 0; i < N; i++)
>     {
>         scanf("%d",&t);
>         if (t < 230348388)
>         {
>             for (j = 0;; j++)
>             {
>                 if (t>a[j] && t <= a[j+1])
>                 {
>                     solve(t, j);
>                     break;
>                 }
>             }
>         }
>         else if (t>=230348388 && t < 944193388)
>         {
>             for (j = 10999;; j++)
>             {
>                 if (t>a[j] && t <= a[j+1])
>                 {
>                     solve(t, j);
>                     break;
>                 }
>             }
>         }
>         else if (t >= 944193388)
>         {
>             for (j = 21999;; j++)
>             {
>                 if (t>a[j] && t <= a[j+1])
>                 {
>                     solve(t, j);
>                     break;
>                 }
>             }
>         }
>     }
>     return 0;
> }
我貌似用得和lz差不多的算法，discuss里的数据也都测过，自己打印大量数据也没发现问题，可就是。。。wa。

• 我貌似用得和lz差不多的算法，discuss里的数据也都测过，自己打印大量数据也没发现问题，可就是。。。wa。 (59) mstchief 2010-02-07 02:45:46