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HDU 5655 CA Loves Stick 水题
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5656 CA Loves Stick Accepts: 381 Submissions: 3204 Time Limit: 2000/1000 MS (Java/Others) Memory Limit ...
分类:其他好文   时间:2016-04-08 21:33:37    阅读次数:209
平衡树---Feed the dogs
POJ 2761 Description Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs every day for Wind. Jiajia loves Wind, b ...
分类:其他好文   时间:2016-04-08 20:00:21    阅读次数:238
hdu5658 CA Loves Palindromic 回文树
回文树在处理回文方面真的比manacher要好用得多。。。 #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #define REP(i,a,b) for(int i ...
分类:其他好文   时间:2016-04-07 07:02:20    阅读次数:191
hdu 5655 CA Loves Stick
CA Loves Stick Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 612 Accepted Submission(s): 214 ...
分类:其他好文   时间:2016-04-04 10:21:14    阅读次数:154
hdu 5656 CA Loves GCD
CA Loves GCD Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 882 Accepted Submission(s): 305 Pr ...
分类:其他好文   时间:2016-04-04 10:19:38    阅读次数:222
数学(GCD,计数原理)HDU 5656 CA Loves GCD
CA Loves GCD Accepts: 135 Submissions: 586 Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) CA Loves GCD Accepts: 13 ...
分类:其他好文   时间:2016-04-04 00:02:05    阅读次数:228
HDU 5656
CA Loves GCD Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 809 Accepted Submission(s): 283 Pr ...
分类:其他好文   时间:2016-04-03 22:03:16    阅读次数:193
HDU 5655 四边形判断
CA Loves Stick Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 563 Accepted Submission(s): 202 ...
分类:其他好文   时间:2016-04-03 21:55:28    阅读次数:150
hdu 5655 CA Loves Stick
Hdu 5655CA Loves Stick 先对四条边排序,若最小的边为0,则输出No(一开始把No和Yes写成了NO和YES,wa了n次) 再判断a[2]+a[1]+a[0]>a[3];满足则能构成四边形,注意此处会爆long long,可以采用减来判断, a[3]-a[2]-a[1]<a[0] ...
分类:其他好文   时间:2016-04-03 14:35:41    阅读次数:152
hdu 5656 CA Loves GCD(dp)
题意: T组数据: n个数字,任取1-n个数字中所有组合的GCD的和 思路:dp /* dp[i][j] 表示用前i个数字gcd=j的所有方法数。 dp[i + 1][gcd(j,a[i + 1])] += dp[i][j]; dp[i + 1][j] += dp[i][j];*/ ...
分类:其他好文   时间:2016-04-03 11:42:28    阅读次数:196
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