
数论
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Farey Sequence - POJ 2478 欧拉公式
Farey SequenceTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11545 Accepted: 4488DescriptionThe Farey Sequence Fn for any integer n with n >= 2 is th原创 2014-04-18 15:43:45 · 909 阅读 · 0 评论 -
Happy 2006 - POJ 2773 欧几里得
Happy 2006Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 10084 Accepted: 3460DescriptionTwo positive integers are said to be relatively prime to each原创 2015-03-31 10:41:47 · 549 阅读 · 0 评论 -
跳蚤 - POJ 1091 容斥原理
跳蚤Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 8723 Accepted: 2601DescriptionZ城市居住着很多只跳蚤。在Z城市周六生活频道有一个娱乐节目。一只跳蚤将被请上一个高空钢丝的正中央。钢丝很长,可以看作是无限长。节目主持人会给原创 2015-03-31 17:22:41 · 1410 阅读 · 0 评论 -
Sky Code - POJ 3904 容斥原理
Sky CodeTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1705 Accepted: 528DescriptionStancu likes space travels but he is a poor software developer an原创 2015-03-31 17:26:46 · 490 阅读 · 0 评论 -
Sky Code - POJ 3904 容斥原理
Sky CodeTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1720 Accepted: 532DescriptionStancu likes space travels but he is a poor software developer an原创 2015-04-03 22:32:58 · 426 阅读 · 0 评论 -
Rectangles - POJ 3695 容斥原理
RectanglesTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 3672 Accepted: 1052DescriptionYou are developing a software for painting rectangles on the s原创 2015-04-04 16:47:04 · 469 阅读 · 0 评论 -
Mike and Foam - CodeForces #547 C 容斥原理
Mike and Foamtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputMike is a bartender at Rico's bar. At Rico's,原创 2015-05-27 22:36:17 · 1102 阅读 · 0 评论 -
Mike and Frog - CodeForces #547 A
Mike and Frogtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputMike has a frog and a flower. His frog is name原创 2015-05-27 14:45:27 · 740 阅读 · 0 评论 -
Bell - HDU 4767 贝尔数
BellTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 600 Accepted Submission(s): 252Problem DescriptionWhat? MMM is learning C原创 2015-07-27 20:59:30 · 572 阅读 · 0 评论 -
Examining the Rooms - HDU 3625 第一类斯特林数
Examining the RoomsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1205 Accepted Submission(s): 727Problem DescriptionA murde原创 2015-07-27 16:50:32 · 426 阅读 · 0 评论 -
Counting Binary Trees - HDU 3240 卡特兰数+逆元
Counting Binary TreesTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 660 Accepted Submission(s): 213Problem DescriptionThere原创 2015-07-14 16:23:12 · 691 阅读 · 0 评论 -
M斐波那契数列 - HDU 4549 费马小定理
M斐波那契数列Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 1990 Accepted Submission(s): 570Problem DescriptionM斐波那契数列F[n]是一种整数数列,原创 2015-07-28 22:17:13 · 593 阅读 · 0 评论 -
SETI - POJ 2065 高斯消元
SETITime Limit: 1000MS Memory Limit: 30000KTotal Submissions: 1698 Accepted: 1056DescriptionFor some years, quite a lot of work has been put into listening to e原创 2015-09-01 22:22:19 · 404 阅读 · 0 评论 -
Widget Factory - POJ 2947 高斯消元
Widget FactoryTime Limit: 7000MS Memory Limit: 65536KTotal Submissions: 5000 Accepted: 1721DescriptionThe widget factory produces several different kinds of wid转载 2015-09-01 20:54:17 · 404 阅读 · 0 评论 -
Flip Game - POJ 1753 高斯消元
Flip GameTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 34598 Accepted: 15136DescriptionFlip game is played on a rectangular 4x4 field with two-sided原创 2015-08-18 22:35:41 · 486 阅读 · 0 评论 -
Polynomial Coefficients - UVa 10105 多项式系数
题意:给你(x1+x2+....+xk)^n的式子,然后给定所有x的次方,问这个它的系数是多少。思路:组合数c(n,k),不用多说。AC代码如下:#include#includeusing namespace std;int n,m,c[20][20],num[20],sum[20],ans;int main(){ int i,j,k; for(i=0;i<=原创 2015-02-16 15:02:38 · 590 阅读 · 0 评论 -
Square root - UVa 10023 大数开方
Square rootThe ProblemYou are to determinate X by given Y, from expression The InputThe first line is the number of test cases, followed by a blank line.Each test case of the input c原创 2015-02-03 16:37:21 · 617 阅读 · 0 评论 -
Visible Lattice Points - POJ 3090 欧拉公式
Visible Lattice PointsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5239 Accepted: 3049DescriptionA lattice point (x, y) in the first quadrant (x原创 2014-04-19 10:36:58 · 898 阅读 · 0 评论 -
Fibonacci - POJ 3070 矩阵乘法快速幂
FibonacciTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8570 Accepted: 6093DescriptionIn the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = F原创 2014-04-24 16:41:53 · 758 阅读 · 0 评论 -
Biorhythms - POJ 1005 中国剩余定理
BiorhythmsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 110898 Accepted: 34511DescriptionSome people believe that there are three cycles in a person原创 2014-05-31 22:43:44 · 721 阅读 · 0 评论 -
Prime Path - POJ 3126 BFS宽度(广度)优先搜索
Prime PathTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10483 Accepted: 5992DescriptionThe ministers of the cabinet were quite upset by the message fr原创 2014-04-15 18:51:01 · 838 阅读 · 1 评论 -
Jzzhu and Sequences - CF 450B 矩阵快速幂版
Jzzhu and Sequencestime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputJzzhu has invented a kind of sequences,原创 2014-07-20 11:36:48 · 914 阅读 · 0 评论 -
So Easy! - HUD 4565 递推→矩阵快速幂
So Easy!Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2089 Accepted Submission(s): 646Problem Description A sequence Sn is原创 2014-07-04 16:44:27 · 908 阅读 · 0 评论 -
Carmichael Numbers - UVa 10006 素数判断
Carmichael NumbersAn important topic nowadays in computer science is cryptography. Some people even think that cryptography is the only important field in computer science, and that life would not原创 2014-12-16 17:57:12 · 619 阅读 · 0 评论 -
Fermat vs. Pythagoras - UVa 106 勾股数
Fermat vs. PythagorasBackgroundComputer generated and assisted proofs and verification occupy a small niche in the realm of Computer Science. The first proof of the four-color problem was co原创 2014-12-18 21:29:46 · 588 阅读 · 0 评论 -
覆盖的面积 - HDU 1255 扫描线 重复两次的面积
覆盖的面积Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3841 Accepted Submission(s): 1887Problem Description给定平面上若干矩形,求出被这些矩形覆盖原创 2014-12-08 16:54:51 · 505 阅读 · 0 评论 -
Base -2 - UVa 11121 -2进制
Base -2 Input: Standard InputOutput: Standard Output The creator of the universe works in mysterious ways. Buthe uses a base ten counting system and likes round numbers.原创 2015-02-02 13:31:20 · 572 阅读 · 0 评论 -
Minimum Sum LCM - UVa 10791
Minimum Sum LCMLCM (Least Common Multiple) of a set of integers is defined as the minimum number, which is a multiple of all integers of that set. It is interesting to note that any positi原创 2015-02-02 14:34:23 · 501 阅读 · 0 评论 -
Send a Table - UVa 10820 欧拉函数
Send a TableInput: Standard InputOutput: Standard Output When participating in programming contests, you sometimes face the following problem: You know how to calcutale the output for the give原创 2015-02-02 18:16:56 · 551 阅读 · 0 评论 -
Leading and Trailing - UVa 11029
Apart from the novice programmers, all others know that you can’t exactly represent numbers raised to some high power. For example, the C function pow(125456, 455) can be represented in double data ty原创 2015-02-03 16:26:24 · 513 阅读 · 0 评论 -
Fox And Jumping - CodeForces 510 D
Fox And Jumpingtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputFox Ciel is playing a game. In this game th原创 2015-02-03 17:33:11 · 724 阅读 · 0 评论 -
Bear and Polynomials 639 C
Bear and Polynomialstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputLimak is a little polar bear. He doesn原创 2016-03-29 21:24:51 · 811 阅读 · 0 评论