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Glenn's ACM code

什麼事都盡全力去做吧!!

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Glenn's ACM code | glennchen1.blogspot.com Reviews
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什麼事都盡全力去做吧!!
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1 glenn's acm code
2 什麼事都盡全力去做吧
3 include stdio h
4 include math h
5 int main
6 if 2q p
7 if 2s r
8 return 0;
9 int i;
10 return tmp;
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Glenn's ACM code | glennchen1.blogspot.com Reviews

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1

Glenn's ACM code: 2009-03-29

http://glennchen1.blogspot.com/2009_03_29_archive.html

Run time: 0.080. Const int N = 1500;. Int ugly[N] = {1, 2, 3, 4, 5};. Int base[3] = {2, 3, 5};. Int min(int *arr);. Int cnt = 5;. For ( j = 0; ugly[j]. Tmp[i] = ugly[j]*base[i];. Index = min(tmp);. Ugly[cnt ] = tmp[index];. Printf("The 1500'th ugly number is %d. n", ugly[N-1]);. Min = arr[0];. Min = arr[1];. 訂閱: 文章 (Atom). C Algorithms [C Reference]. A Tutorial on Dynamic Programming. Lucky 貓的 ACM 園地. Awesome Inc.範本. 由 Blogger.

2

Glenn's ACM code: 2009-03-08

http://glennchen1.blogspot.com/2009_03_08_archive.html

Q10220: I Love Big Numbers! Run time: 0.030. Int len[N 1];. While (scanf("%d", &n)! Printf("%d n", len[n]);. Arr[0] = 1;. Len[0] = 1, len[1] = 1;. For ( j = 0; j LEN; j ). Arr[j] *= i;. For ( j = 0; j LEN; j ). If (arr[j] = 10). Arr[j 1] = arr[j]/10;. Arr[j] %= 10;. For ( j = 0; j LEN; j ). Len[i] = arr[j];. 上傳時再改成 LEN: 2600, N: 1000. Run time: 0.410. Int arr[N 1][LEN];. While (scanf("%d", &n)! For ( i = LEN-1; i = 0; i- ). Printf("%d", arr[n][i]);. Arr[0][0] = 1, arr[1][0] = 1;. For ( j = 0; j LEN; j ).

3

Glenn's ACM code: 2009-02-15

http://glennchen1.blogspot.com/2009_02_15_archive.html

Run time: 0.010. Int num; / 輸入組數. Int i; / for loop count. Double s, e; / a(0) and a(n 1). Double sum, total;. Scanf("%d", &num);. Sum = total = weight = 0;. Scanf("%d", &n);. Scanf("%lf%lf", &s, &e);. Scanf("%lf", &arr[i]);. Sum = s*n e;. For ( i = n-1; i = 0; i- ). Total = (weight =2)*arr[i];. Printf("%.2lf n", (sum-total)/(n 1) ;. Run time: 0.000. Char str[N][N 1];. Char arr[N 1];. Void check(int length, char ch, int row, int len);. Bool test(int row, int col, int length, int row, int len);. If (c-len...

4

Glenn's ACM code: 2009-02-08

http://glennchen1.blogspot.com/2009_02_08_archive.html

Q343: What Base Is This? Run time: 0.010. Int find(int *num, int len);. Int calculate(int *num, int len, int base);. Char str1[100], str2[100];. Int arr1[100], arr2[100];. Int len1, len2;. Int i; / base1. Int j; / base2. Int m1, m2;. While (scanf("%s%s", str1, str2) = 2). Len1 = strlen(str1);. Len2 = strlen(str2);. For ( i = 0; i len1; i ). If (str1[i] = '0' & str1[i] = '9'). Arr1[i] = int(str1[i])-48;. Arr1[i] = str1[i]- 'A' 10;. For ( i = 0; i len2; i ). If (str2[i] = '0' & str2[i] = '9'). If (x = y).

5

Glenn's ACM code: 2009-02-22

http://glennchen1.blogspot.com/2009_02_22_archive.html

Contestant problem time L. 其中contestant為參賽者編號(1~100). Problem為題目編號(1~9). L為裁判結果,分為C、I、R、U、E。C代表正確,I代表不正確,後三者不影響分數. Run time: 0.000. Void assign(int n, int question, int time, char result);. Scanf("%d n", &n);. If (str[0] = ' 0') break;. Sscanf(str, "%d %d %d %c", &num, &q, &t, &ch);. Assign(num, q, t, ch);. Calculate the penalty: (number of wrong answer)*20 accepting time. 1 solve the number of problems. 2 the less penalty, the better rank. Record[i].prob[j] = false;. Record[i].wrong[j] = 0;. S[i] = true;.

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One day at a time. The Pumphouse Restaurant Dinner Buffet May 2, 1984. Fairbanks, Alaska. Glenn Moehlin. Oyster Bar Chef in 1984 at The Pumphouse Restaurant in Fairbanks Alaska. Aloha! Ldquo;Trust in the Lord with all your heart,. And lean not on your own understanding;. In all your ways acknowledge him,. And he will make your paths straight.”. Don't Drink and Drive! The life you save may be your own. Share this page on Facebook.

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Glenn's ACM code

Q10375: Choose and divide. Exp(x) return a double, so we have to convert its datatype to long double. Long double: printf("%Lf"), scanf("%LF"). Run time: 0.056. Long double compute(int m, int n);. Long double tmp1, tmp2, result;. While (scanf("%d%d%d%d", &p, &q, &r, &s) = 4). Q) tmp1 = 0;. Else tmp1 = compute(p, q);. S) tmp2 = 0;. Else tmp2 = compute(r, s);. Result = exp(tmp1-tmp2);. Printf("%.5Lf n", result);. Long double compute(int m, int n). Long double tmp = 0;. Tmp = log(i);. Tmp -= log(i);.

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