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Common BoardI've WA#8 too.Interesting, what test is it?Thank!!! Edited by author 27.10.2007 17:54 And I had WA too; because of stupid mistakes. This tests helped me to figure them out: 5 0 0 0 2 1 1 2 0 2 2 9 0 0 0 10 0 20 10 0 10 10 10 20 20 0 20 10 20 20 The thing is that if you order edges by an angle (atan) and length and do binsearch it's still possible that there will be edges with an appropriate angle, bigger length but not the one you are looking for, so binsearch in this case will give you a wrong result, so try to find a way to order them simple test is 6 0 0 10 0 10 10 10 20 0 10 0 20 and the answer is 171 #include <stdio.h> int main () { int n, m; scanf("%d %d", &n, &m); int ara[n+1]{0}, temp; double result{0}; for (int i = 1; i <= m; i++) { scanf("%d", &temp); ara[temp]++; temp = 0; } for (int i = 1; i <= n; i++) { if(m == 0) { printf("0.00%\n"); continue; } result = (100.0 * ara[i] / m); printf("%2.2lf", result); printf("%\n"); } return 0; } What is test nubmer 4? Anybody know? maybe 6 1 1 1 1 1 1 My program can`t pass this test and i have WA4. It's like this: 15 2 5 6 3 8 2 3 6 10 10 12 13 14 15 11 The right answer is: 5 8 Edited by author 24.01.2010 23:08 No the right answer is 3 8 for (long double N = 1; N < sqrt(2*S)+3; N++) { } Я решил с помощью ДП: f[i,k,d]-число способов выбрать k чисел от i до s, нод которых d. Работает 0.031 sec. Вопрос: как решить эту задачу именно с помощью теории чисел( я использовал в динамике только один факт: gcd(a1,a2,...,ak)=gcd(a1,gcd(a2,...,ak)))? Актуально ли это) но здесь можно решать еще с помощью формулуы включений-исключений. Предподсчитаем простые числа. Признаки в формуле - делимость на i. сочетаниями подсчитаем кол-во способов выбрать наборы. сочетания посчитаем динамикой I have problems with english. Sorry for example you answer is = int(ans) but use int(ans + 0.00001) and you will get AC There are only 6 available combinations. Take a dice and think if you can make it so that 1 is adjacent to 6? 2 to 5? 3 to 4? Test case: 5 5 Ans: 5 1 3 4 2 Test case: 10 2 Ans: 2 4 6 8 1 5 9 7 3 Is it correct ans? Of course 2nd one is wrong, it skips 10. I think, the correct answer for your second test is: 2 4 6 8 10 3 7 1 9 5 Can anybody tell me test 10? "Impossible" should start with uppercase 'I' 参考 百度知道 给定L,M,K,N,求序列A的数量。满足: 1. A的长度为L 2. A中元素均在 [1,M] 间的正整数 3. A的元素之和为K的倍数 求字典序第N小的序列A,排名从0开始标号 (即,称最小为“第0小”) L<=100,M<=50,K<=50 N范围题目没有给 Edited by author 26.12.2021 13:22 Edited by author 26.12.2021 13:22 Can you help me with test 28? Yea, sure. Just try 6 6 ****** ****** **..** **..** ****** ****** answer: 1 Anyone has any tests for WA #6? I am out of ideas already. The problem seems too simple... This test helps you: 10 1 2 3 4 5 6 7 8 9 0 4 3 answer: 39 6 The problem says the program only needs to print out any of the answers. but why i WA in the first test but can pass all my tests? If you are storing hashes in a segtree and using some custom prime divider the values of the parent nodes can be less than the child nodes because the remainder "makes a round", so to subtract u have to add that round, I think this is what is actually happening in test 2. Subj. Btw, what is the problem with the 23rd test? it's help me for wa23 4 0 4 0 4 0 1 0 1 1 0 1 1000000 0 1000000 0 Indeed! Thank you very much! Thank you very much! it's realy helpful!!! Very help!! Thank you again! Help please!! What is test 6???? Just try to replace x, y or z using equation y = x + z (vector form). Count of cells on path is |x| + |y| + |z| + 1. This is brilliant Just try to replace x, y or z using equation y = x + z (vector form). Count of cells on path is |x| + |y| + |z| + 1. Testset for this problem is very weak. Program, that works ~5 secs. on my tests got AC in less than 1 sec. on jury tests... why don't you share your strong tests with us then? |
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