摘要:数学归纳法,得证只需求得使18+ka被64整除的a。且a不超过65。#include int main() { int i, j, k; while (scanf("%d", &k) != EOF) { j = 0; for (i=0; i<65; i++) { if ((18+k*i) % 65 == 0) { j = 1; break; } } if (j) printf("%d\n", i); ...
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摘要:水题,如题。#include #define chk(Y) (Y%4==0 && Y%100!=0) || Y%400==0int main() { int case_n; int i, y, n; scanf("%d", &case_n); while (case_n--) { scanf("%d %d", &y, &n); i = 0; while (1) { if ( chk(y) ) i++; if (i) { if ...
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摘要:水题,原理是计算机组成原理中的负数的补码的求码。利用按位与可解。#include using namespace std;int main() { int n; while (1) { cin >> n; if (n == 0) break; cout <<(n&(-n))<<endl; } return 0;}
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摘要:水题,暴力可解。#include using namespace std;int chg(int n, int base);int main() { int i; int tmp; for (i=2992; i 0) { sum += n % base; n = n / base; } return sum;}
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摘要:此题主要采用DP思路,难点是求解"/",需要考虑划分数量不够的情况,先采用DP求解最优解,然后采用贪心求解slash。防止中间结果出错,使用了unsigned int。#include using namespace std;#define MAXNUM 501unsigned int pages[MAXNUM][MAXNUM];int main() { int case_n, m, k; int i, j, p, q; unsigned min, tmp, sum; cin >>case_n; for (p=1; p>m>>k; for (i
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摘要:最开始使用递归DP解,stack overflow。化简了一些,复杂度为O(n)就过了。#include int main() { int case_n, n; int i, j, tmp; int beg, end, sum, max_sum; int number; scanf("%d", &case_n); for (i=1; i max_sum) { max_sum = sum; beg = tmp; end = j; } ...
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摘要:经典DP,写的可能麻烦了一些。#include #define false 0#define true 1int is_prime[41];int is_visit[21];int sequence[21];void DFS(int, int, int);int main() { int i, j, k; int n; is_prime[1] = false; is_prime[2] = true; for (i=2; i<=40; i++) { k = 0; for (j=2; j*j<=i; j++) { i...
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摘要:肯定是大叔,本来以为用加法做乘法,后来想想这样麻烦,还是可以使用乘法的,按位乘,ov可以看成不止一位,程序如下所示:#include #define MAX 40000int main() { unsigned int res[MAX]; int n; int ov, i, j, k; int pos; while (scanf("%d", &n) != EOF) { res[0] = 1; pos = 1; for (i=1; i=0; i--) printf("%d", res[i]); ...
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摘要:题目主要找发声类似于Peak Hell的Python模块,查了一下手册pickle已经是最像的了。看了一下源代码,发现panner.p。如同发现了新大陆,拷贝内容。使用pickle解答。答案为channel。程序如下:import picklefilein = open("E:\\Python\\data", "rb")lst = pickle.load(filein)print(type(lst)) # Find It is a list, You'd better saw the content first.string = "&q
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摘要:题目为正则表达式,需要注意EXACTLY的含义,即AAAAxBBB中x不满足条件,但aAAAxBBBa却满足条件。使用perl解决此题,利用s///替换字母,循环读取整个源码文件,结果为linkedlist。解答完毕,Perl程序如下:#!perl -wuse 5.010;############## Level 3 ############### data store the sourcepage informationopen FILEIN, "data" or die "Can't open the data."; my $word = &
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