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A|B? Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18605 Accepted Submission(s): 14122 Problem 阅读全文
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Max Num Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 22270 Accepted Submission(s): 13029 Problem Description There are some students in a ... 阅读全文
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Fibbonacci Number Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 22530 Accepted Submission(s): 10375 Problem Description Your objective for ... 阅读全文
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#include using namespace std; long long dp[36][36]; int main() { int num; int i = 1; for (int i = 0; i > num && num != -1) { printf("%d %d %I64d\n", i++, num, 2 * ... 阅读全文
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超时 a1 = k an = k + n - 1 M = n*(2k+n-1)/2 解的 k = M/n - (n-1)/2，这个不能这样写，要写在一起 (2M-n*n-n)/(2*n)。否则对于30/4 - 3/2，会忽略这样的结果，如果先通分当然也就可以。 超时 利用a1>=1这个条件可以减少循 阅读全文
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#include #include #include using namespace std; /* 无穷多次操作，第n灯的状态 */ int main() { long long n; while (cin >> n) { int count = 0; for (int i = 1; i <= n; i++) ... 阅读全文
摘要：
#include #include using namespace std; int main() { int n, m; while (cin >> m >> n) { for (int i = 0; i <= n+1; i++) { if (i == 0 || i == n+1) ... 阅读全文
