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C. Number Transformation II time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output C. Number Tra 阅读全文
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Queue-jumpers Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3348 Accepted Submission(s): 904 P 阅读全文
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No Pain No Game Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2000 Accepted Submission(s): 851 阅读全文
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Group Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1959 Accepted Submission(s): 1006 Problem 阅读全文
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Sequence operation Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7502 Accepted Submission(s): 阅读全文
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Transformation Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 65535/65536 K (Java/Others) Total Submission(s): 4095 Accepted Submission(s): 100 阅读全文
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威威猫系列故事——晒被子 Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 1592 Accepted Submission(s): 444 Pr 阅读全文
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Memory Control Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5913 Accepted Submission(s): 1380 阅读全文
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A Task Process Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1332 Accepted Submission(s): 656 阅读全文
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（1）Lucas定理：p为素数，则有: （2）证明： n=(ak...a2,a1,a0)p = (ak...a2,a1)p*p + a0 = [n/p]*p+a0，m=[m/p]*p+b0其次，我们知道，对任意质数p有(1+x)^p=1+(x^p)(mod p) 。我们只要证明这个式子：C(n,m) 阅读全文
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J. Ceizenpok’s formula time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output J. Ceizenpok’s fo 阅读全文
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Segment Accepts: 418 Submissions: 2020 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) 问题描述 \ \ \ \ Rivendell非常神,喜欢研究 阅读全文