hd 1597 find the nth digit(解决超时的牛气哄哄的新办法)

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<h1 style="COLOR: #1a5cc8"><div class="panel_title" align="left"><h1 style="COLOR: #1a5cc8">find the nth digit</h1></div><span size="+0"><strong><span style="font-family:Arial;font-size:12px;color:green;FONT-WEIGHT: bold">Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8385    Accepted Submission(s): 2394</span></strong></span></h1><div class="panel_title" align="left">Problem Description</div><div class="panel_content">假设：S1 = 1S2 = 12S3 = 123S4 = 1234.........S9 = 123456789S10 = 1234567891S11 = 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？</div><div class="panel_bottom"> </div><div class="panel_title" align="left">Input</div><div class="panel_content">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)。</div><div class="panel_bottom"> </div><div class="panel_title" align="left">Output</div><div class="panel_content">对于每个N，输出S中第N个对应的数字.</div><div class="panel_bottom"> </div><div class="panel_title" align="left">Sample Input</div><div class="panel_content"><pre><div style="FONT-FAMILY: Courier New,Courier,monospace">61234510</div>

Sample Output

Author

8600

Source

HDU 2007-Spring Programming Contest - Warm Up （1）

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#include<stdio.h>//#include<iostream>#include<math.h>//using namespace std;int main(){    __int64 n,t,i,s,p;    scanf("%d",&t);     while(t--)     {       scanf("%I64d",&n);       for(i=(int)sqrt(n*2)-1;;i++)       {       if(i*(i+1)/2>=n)       {        p=i;        break;                      }                      }       s=(p-1)*p/2;       if(n>s)       n-=s;       while(n>9)       {       n-=9;       }       printf("%d\n",n);                              }return 0;    }//加上限制条件i<n试试，不要想当然，多思 //注意初始限制条件i=sqrt(n*2)-1;//超时时如果确认过程正确的话试着把iint 改__int64 下，所有的变量都改着试试

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