hd 1597 find the nth digit(解決超時的牛氣哄哄的新辦法)
阿新 • • 發佈：2019-02-02
<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">假設： S1 = 1 S2 = 12 S3 = 123 S4 = 1234 ......... S9 = 123456789 S10 = 1234567891 S11 = 12345678912 ............ S18 = 123456789123456789 .................. 現在我們把所有的串連線起來 S = 1121231234.......123456789123456789112345678912......... 那麼你能告訴我在S串中的第N個數字是多少嗎？ </div><div class="panel_bottom"> </div> <div class="panel_title" align="left">Input</div><div class="panel_content">輸入首先是一個數字K，代表有K次詢問。 接下來的K行每行有一個整數N(1 <= N < 2^31)。</div><div class="panel_bottom"> </div> <div class="panel_title" align="left">Output</div><div class="panel_content">對於每個N，輸出S中第N個對應的數字. </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">6 1 2 3 4 5 10</div>
Sample Output 1 1 2 1 2 4
Author 8600
Source
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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; } //加上限制條件i<n試試，不要想當然，多思 //注意初始限制條件i=sqrt(n*2)-1; //超時時如果確認過程正確的話試著把iint 改__int64 下，所有的變數都改著試試