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HDU4421 Bit Magic 【2-sat】
叙述性说明:这给出了一个矩阵,原来的请求a排列2-sat称号。对于每一位跑步边,跑31位可详细的施工方注意N=1的情况特判,还有检查对称元素是否同样#include #include #include #include #include #include #define pb push_back#i...
分类:其他好文   时间:2015-07-25 21:21:16    阅读次数:130
Problem A CodeForces 560A
DescriptionA magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the sys...
分类:其他好文   时间:2015-07-25 21:20:20    阅读次数:132
Currency System in Geraldion
standard outputA magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the...
分类:其他好文   时间:2015-07-25 19:45:58    阅读次数:111
Currency System in Geraldion (Codeforces 560A)
A Currency System in GeraldionTime Limit:2000MSMemory Limit:262144KB64bit IO Format:%I64d & %I64uDescriptionA magic island Geraldion, where Gerald li....
分类:其他好文   时间:2015-07-25 16:43:25    阅读次数:123
Codeforces Round #313 A Currency System in Geraldion
A Currency System in GeraldionTime Limit:2000MSMemory Limit:262144KB64bit IO Format:%I64d & %I64uDescriptionA magic island Geraldion, where Ger...
分类:其他好文   时间:2015-07-25 15:11:39    阅读次数:104
ACM判断所给数字能否表示所有的数字
DescriptionA magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the sys...
分类:其他好文   时间:2015-07-25 15:04:24    阅读次数:524
Currency System in Geraldion
题目:DescriptionA magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the ...
分类:其他好文   时间:2015-07-25 15:01:17    阅读次数:118
货币问题
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64uDescriptionA magic island Geraldion, where Gerald lives, has its own currency sys...
分类:其他好文   时间:2015-07-24 22:05:51    阅读次数:89
Codeforces Round #313 A. Currency System in Geraldion
Description A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens th...
分类:其他好文   时间:2015-07-24 20:52:47    阅读次数:101
SGU Magic Pairs
A0x + B0y = knAx + By = k'n左差得(A - A0)x + (B -B0)y = 0(mod n)所以只要枚举A0, B0的倍数就行了。。公式就是 ( (i*a)%n, (i*b)%n ), i =0, 1, ... , n-1i*a, i*b如果大于n的话 不会影响结果, ...
分类:其他好文   时间:2015-07-24 17:35:55    阅读次数:87
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