1 / 21

10624 - Super Number

10624 - Super Number. 99703005 資科二 謝濟謙 99703030 資科二 薛元昊 99703038 資科二 鄭凱原. Outline. Problem Analyze Solution. Problem - Super Number.

aelan
Download Presentation

10624 - Super Number

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 10624 - Super Number 99703005 資科二 謝濟謙 99703030 資科二 薛元昊 99703038 資科二 鄭凱原

  2. Outline • Problem • Analyze • Solution

  3. Problem - Super Number • Don't you think 162456723 very special? Look at the picture below if you are unable to find its speciality. (a | b means ‘b is divisible by a’) • Figure: Super Numbers

  4. Problem - Super Number • Input: Standard Input • Output: Standard Output • Time Limit: 3 Seconds • Givenn, m (0 < n < m < 30), you are to find a m-digit positive integer X such that for every i (n <= i <= m), the first i digits of X is a multiple of i. If more than one such X exists, you should output the lexicographically smallest one. Note that the first digit of X should not be 0.

  5. Problem - Super Number • Input  • The first line of the input contains a single integer t(1 <= t <= 15), the number of test cases followed. For each case, two integers n and m are separated by a single space. • Output  • For each test case, print the case number and X. If no such number, print -1. • Sample Input                           Output for Sample Input

  6. Analyze • Example: • Input : 1 5 15 • Output:

  7. Analyze • 大數運算: • 基於記憶體的有效運用，程式語言中規定了各種不同的資料型態，也因此變數所可以表達的最大整數受到限制，例如123456789123456789這樣的 整數就不可能儲存在long變數中（例如C/C++等），我們稱這為long數，這邊翻為超長整數（避免與資料型態的長整數翻譯混淆），或俗稱大數運算。

  8. Analyze • 就C語言來看，long longint已經是最大的整數型態了！ • (long longint型態的數值，範圍是：-264-1~ 264-1-1)

  9. Analyze 這個是長除法，而以下所解釋的就是 ”大數除法”的實現

  10. Analyze current depth x = 0 remain = ( remain * 10 + a[x] ) % Divisor =>( 0 * 10 + 3 ) % 195 = 3 remain = 3

  11. Analyze current depth x = 1 =>( 3 * 10 + 0 ) % 195 = 30 remain = 30 current depth x = 2 =>( 30 * 10 + 1 ) % 195 = 106 remain = 106 current depth x = 3 =>( 106 * 10 + 0 ) % 195 = 85 remain = 85

  12. Analyze current depth x = 4 remain = 74 =>( 85 * 10 + 4 ) % 195 = 74 current depth x = 5 remain = 163 =>( 74 * 10 + 8 ) % 195 = 163 Answer is 163

  13. Analyze 1. intmain(void) { 2. int times = 0, count; 3. scanf(“%d”, &times); //輸入幾筆測資 4. for (count = 0; count < times; ++count) { 5. scanf(“%d %d”, &n, &m); // 輸入起始數字的總位數與結束數字的總位數 6. printf(“Case %d: ”, count + 1); //當該筆測資之答案找到時輸出 7. if(superNum(0)) { // 8. for (inti = 0; i < m; ++i) // 9. printf(“%d”, superNumber[i]); // 依位數依序印出答案 10. printf("\n"); // 11. } // 12. else 13. printf(“-1\n”); // 若該筆測資沒有符合的答案，則輸出-1 14. } 15. return 0; // 程式結束 16. }

  14. Analyze 1. boolbigNumberMod(int depth) { 2. int remain = 0; 3. for (inti = 0; i < depth; ++i) 4. remain = (remain * 10 + superNumber[i]) % depth;// 引用大數運算方法計算 5. return !remain; • } 8. boolsuperNum(int depth) { 9. if (depth == m) 10. return true; • for (inti = 0; i < 10; ++i) { • if (depth == 0 && I == 0) ++i; 13. superNumber[depth] = i; 14. if (((depth < (n - 1)) || bigNumberMod(depth + 1)) && superNum(depth + 1)) //遞迴方法 15. return true; 16. } 17. return false; 18. }

  15. Better Solution #include<stdio.h> int n, m, superNumber[30]; boolbigNumberMod(int depth) { int remain = 0; for (inti = 0; i < depth; ++i) remain = (remain * 10 + superNumber[i]) % depth; return !remain; } boolsuperNum(int depth) { if (depth == m) return true; for (inti = 0; i < 10; ++i) { superNumber[depth] = i; if (((depth < (n - 1)) || bigNumberMod(depth + 1)) && superNum(depth + 1)) return true; } return false; }

  16. Better Solution int main(void) { int times = 0, count; scanf("%d", &times); for (count = 0; count < times; ++count) { bool found = false; scanf("%d %d", &n, &m); for (inti = 1; i < 10; ++i) { superNumber[0] = i; if (superNum(1)) {found = true; break;} } printf("Case %d: ", count + 1); if(found) { for (inti = 0; i < m; ++i) printf("%d", superNumber[i]); printf("\n"); } else printf("-1\n"); } return 0; }

  17. Solution • Time complexipy • T = O(m^2) Approaching to TLE? ↑ (3s)

  18. Solution Approaching to TLE ↑ (3s)

  19. Solution <Data Base> Int main(){ int times, m, n, i; scanf("%d", &times); for( i = 1 TO times ){ scanf("%d %d", &n, &m); cout<<"Case "<<i<<": "; if( m == 29 ){ if( n < 6 )cout<<“-1”<<endl; else if( n == 7 )cout<<“out29[1]”<<endl; else if( n == 8 To 28 ) … } else if( m == 28 To 2 ) … } return 0; } string out29[23] = {"39208896001212006680167258804","18555536703606084000060008482","10211576408418088880443254681","10101279606082569080443254680","10015192000842085860660000924","10000351372830084060967250126","10000028419290081060024006085","10000003819134081660600004206","10000000299310567240867250403","10000000177423525880167258806","10000000012965287440500008129","10000000000552924880760008767","10000000000010730620467258241","10000000000000635540300008128","10000000000000024060207252565","10000000000000000485667250123","10000000000000000135423252801","10000000000000000007244006366","10000000000000000001503258209","10000000000000000000006750880","10000000000000000000000224923","10000000000000000000000141200","10000000000000000000000000086"};

  20. Solution <Data Base> • Time complexipy • T = O(1) 2.960/0.008 = 370

  21. End Thanks for your listening!

More Related