题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1597

find the nth digit

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9517    Accepted Submission(s): 2757

Problem Description假设：
S1 = 1
S2 = 12
S3 = 123
S4 = 1234
………
S9 = 123456789
S10 = 1234567891
S11 = 12345678912
…………
S18 = 123456789123456789
………………
现在我们把所有的串连接起来
S = 1121231234…….123456789123456789112345678912………
那么你能告诉我在S串中的第N个数字是多少吗？ Input输入首先是一个数字K，代表有K次询问。
接下来的K行每行有一个整数N(1 <= N < 2^31)。 Output对于每个N，输出S中第N个对应的数字. Sample Input6
1
2
3
4
5
10 Sample Output1
1
2
1
2
4 Author8600题解：可以用二分解，也可以用二次方程公式解，要注意使用long long二分查找的代码：

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 using namespace std;
 #define ll long long
 ll f(ll x)
 {
     ll ans = x*(x+)/ ;
     return ans;
 }
 int main()
 {
     int T;
     scanf("%d",&T);
     for(int i = ;i < T ;i++)
     {
         ll n ;
         scanf("%lld",&n);
         if(n==)
         {
             puts("");
             continue;
         }
         ll l = ;
         ll r = n;
         ll mid =( l+r )/;
         int tm = ;
         while(r-l>)
         {
             if( f(mid) < n )
             {
                 l = mid;
                 mid = (l+r)/;
             }
             else if( f(mid) > n)
             {
                 r = mid;
                 mid = (l+r)/;
             }
             else if( f(mid)==n )
             {
                 tm = mid;
                 break;
             }
         }
         if(tm!=)
         {
             if(tm%==) puts("");
             else
             printf("%d\n",tm%);
         }
         else
         {
             ll flag = l;
             ll tt = flag*(flag+)/;
             int sum = n-tt;
             if(sum%==) puts("");
             else
             printf("%d\n",sum%);
         }
     }
     return ;
 }

公式代码：

 #include<cstdio>
 #include<cmath>
 using namespace std;
 #define ll long long
 #define eps 1e-8 int main()
 {
     int T;
     scanf("%d",&T);
     for(int i = ;i < T;i++)
     {
         double n;
         scanf("%lf",&n);
         double m =(sqrt(+*n)-)/;
         ll mm =m;
         ll tm = mm*(mm+)/;
         ll flag = n-tm;
         if(flag == ){if(mm%==)printf("9\n");else   printf("%lld\n",mm%);}
         else
         {
             if(flag%==) printf("9\n");
             else
             printf("%lld\n",flag%);
         }
     }
     return  ;
 }