随笔分类 - 简单数论/数学/数学思想
Educational Codeforces Round 8 B 找规律
摘要:B. New Skateboard time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output B. New Skateboard time
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Manthan, Codefest 16 B 数学
摘要:B. A Trivial Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output B. A Trivial Proble
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Codeforces Round #240 (Div. 2) B 好题
摘要:B. Mashmokh and Tokens time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output B. Mashmokh and To
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Codeforces Round #320 (Div. 2) [Bayan Thanks-Round] D 数学+(前缀 后缀 预处理)
摘要:D. "Or" Game time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output D. "Or" Game time limit per
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Codeforces Beta Round #95 (Div. 2) C 组合数学
摘要:C. The World is a Theatre time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output C. The World i
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HDU 4611
摘要:Balls Rearrangement Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 2862 Accepted Submission(s):
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Codeforces Round #324 (Div. 2) D
摘要:D. Dima and Lisa time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output D. Dima and Lisa time li
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ACdream 1023 抑或
摘要:Xor Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) Submit Statistic Next Problem Problem Description For given multi
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2016"百度之星" - 资格赛(Astar Round1) A
摘要:Problem A Accepts: 1551 Submissions: 11043 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Problem A Accepts: 1551 Su
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2016华中农业大学预赛 E 想法题
摘要:Problem E: Balance Description Every night Diao Ze is dreaming about the gold medal in the <!--?mso-application progid="Word.Document"?--> 1540th"> AC
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2016华中农业大学预赛 B 数学
摘要:Problem B: Handing Out Candies Description After the 40th ACM-ICPC, Diao Yang is thinking about finding a girlfriend because he feels very lonely when
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费马小定理求逆元 以及求逆元普遍式子总结
摘要:费马小定理(Fermat Theory) 假如p是质数,且(a,p)=1,那么 a(p-1)≡1(mod p)。即:假如a是整数,p是质数,且a,p互质(即两者只有一个公约数1),那么a的(p-1)次方除以p的余数恒等于1。 当涉及取模运算的计算中,如果有除法,不能直接除以一个数,而应该变成乘以它的
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Codeforces Round #326 (Div. 2) B Duff in Love 简单数论 姿势涨
摘要:B. Duff in Lovetime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputDuff is in love with lovely numbe...
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