
joj acm_大数运算
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2156: Drunk
2156: DrunkStatusIn/OutTIME LimitMEMORY LimitSubmit TimesSolved UsersJUDGE TYPEstdin/stdout3s8192K1158262StandardJack is drunk again and he wants to walk out of原创 2010-03-12 20:21:00 · 426 阅读 · 0 评论 -
1267: Pascal's Triangle of Death
<br /> 1267: Pascal's Triangle of DeathResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K35889Standard<br />In this problem, you are asked to generate Pascal's Triangle. Pascal's Triangle is useful in many areas from probability to polynomials to pr原创 2010-07-21 09:55:00 · 781 阅读 · 0 评论 -
2506: Pseudoprime numbers
<br /> 2506: Pseudoprime numbersResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s16384K26572Standard<br />Fermat's theorem states that for any prime number p and for any integer a > 1, a^p == a (mod p). That is, if we raise a to the pth power and divide原创 2010-07-10 19:55:00 · 589 阅读 · 0 评论 -
1600: Big Mod
<br />1600: Big ModResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE3s8192K1658309Standard<br />Calculate <br /> <br /><br /> <br />for large values of B, P, and M using an efficient algorithm. (That's right, this problem has a time dependency !!!.) <br /原创 2010-07-10 19:57:00 · 594 阅读 · 0 评论 -
1996: Super Addition
<br />1996: Super AdditionResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE60s10240K2026141Standard<br />The sum of any two integers A=a1a2...an and B=b1b2...bn is very easy to make(of course ai or bi may be zero), especially to programmers, because they原创 2010-07-10 19:54:00 · 480 阅读 · 0 评论 -
1199: Just the Facts 求N!的非零末位
1199: Just the FactsStatusIn/OutTIME LimitMEMORY LimitSubmit TimesSolved UsersJUDGE TYPEstdin/stdout3s8192K734332StandardThe expression N!, read as ``N factoria原创 2010-03-13 13:25:00 · 382 阅读 · 0 评论 -
1189: Integer Inquiry
1189: Integer InquiryStatusIn/OutTIME LimitMEMORY LimitSubmit TimesSolved UsersJUDGE TYPEstdin/stdout1s8192K879320StandardOne of the first users of BITs new supe原创 2010-03-13 13:23:00 · 349 阅读 · 0 评论 -
大整数的阶乘的末尾有多少的0
#includeint main(){ int nn,n; scanf("%d",&nn); while(nn--) { scanf("%d",&n); int count=0; while(n) { count+=n/5; n/=5; } printf("%d/n",count); } return 0;}/*算一个大整数的阶乘的末尾有多少的0.算法:不断的将该数除以5,原创 2010-03-12 20:29:00 · 564 阅读 · 0 评论 -
1920: Jojer 大树模板
1920: JojerStatusIn/OutTIME LimitMEMORY LimitSubmit TimesSolved UsersJUDGE TYPEstdin/stdout3s8192K652204StandardMr Jojer is a teacher, and he is crazy in progr原创 2010-03-12 20:26:00 · 1008 阅读 · 0 评论 -
1949: Fibonacci Numbers
1949: Fibonacci NumbersStatusIn/OutTIME LimitMEMORY LimitSubmit TimesSolved UsersJUDGE TYPEstdin/stdout3s16384K775220Special TestA Fibonacci sequence is calculat原创 2010-03-12 20:22:00 · 723 阅读 · 0 评论 -
hdu 1005 Number Sequence 矩阵连乘
Number Sequence<br />Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)<br />Total Submission(s): 37463 Accepted Submission(s): 7885<br /><br />Problem DescriptionA number sequence is defined as follows:<br /><br />f(1)原创 2011-02-21 19:55:00 · 690 阅读 · 0 评论