打表

        写这篇文章是告诉大家我的一个打表方法，开始的时候自己觉得打一个1000的表比较困难，但是真的自己去尝试的话不要几分钟~~什么都得自己去尝试，当然下面这道题肯定有比打表更加适合的方法。

       zjut 1218(n!中的0)题目链接：http://acm.zjut.edu.cn/ShowProblem.aspx?ShowID=1218

      与以前求阶乘中的0的个数不同的是以前求n!中的0是只需求尾数的0即可，希望各位看清楚题目。不会做，没有找到规律，但是知道求高精度的阶乘值，但是超时，没办法，想到了打表，开始时候觉的好复杂，但实际操作时觉得不是那么复杂，主要是介绍下我的打表过程：

      题目要求是1000个数字，我先写了一个生成1-1000的算法，for循环，并记录在新建记事本中，并复制到高精度模板中，生成每个对应的值，按全选后复制运行结果，然后写个打表程序，具体代码为：

#include<iostream>using namespace std;int a[200];//因为在dev的条件下可能运行很多组数据后只会出现后面点的数据,自己可以改动int main(){ int i,n; for(i=1;i<=200;i++) cin>>a[i]; for(i=2;i<=200;i+=2) cout<<a[i]<<','; return 0;}

后将前面复制的数据放到高精度阶乘的模板(怎样在DEV中复制粘贴----对于有很多组测试数据时，我们最好的方法是借助记事本，先将要测试的数据写入到记事本，然后调试的时候右键单击调试框上的黑色部分，在下拉列表中选中粘贴，要提取数据的话可以全选后复制即可)中调试，就会直接出现我们所需的数据排列，后复制到我们的打表程序中即可，所以打一个从1-1000的表并不要几分钟。
高精度阶乘模板：

#include<iostream>#include<cstring> using namespace std;#define max 10000string s1;int s[max];int main(){ int i,j,n,sum; while(cin>>n) { memset(s,0,sizeof(s)); s[0]=1; for(i=2;i<=n;i++) {int c=0; for(j=0;j<max;j++) { sum=s[j]*i+c; s[j]=sum%10; c=sum/10; } } s1=""; for(j=max-1;j>=0;j--)if(s[j])break; for(i=j;i>=0;i--) s1+=s[i]+'0'; //cout<<s1<<endl; //高精度阶乘的值 int num=0; for(i=0;i<s1.length();i++) if(s1[i]=='0') num++; cout<<num<<endl; } return 0;} 

下面是这道题的打表程序：

 #include<iostream>using namespace std;int a[1001]={ 0,0,0,0,0,1,1,2,2,1,2,2,4,4,2,4,4,4,5,6,7,7,8,5,6,9,8,9,10,7,9,7,10,8,11,9,10, 12,16,12,9,15,13,13,12,13,16,11,14,14,19,18,18,17,18,18,17,20,17,19,19,26,20, 21,20,20,23,22,25,21,20,25,23,35,23,25,24,32,25,29,27,27,28,30,27,36,28,27,28, 34,28,29,29,32,37,32,36,26,35,28,30, //100 35,37,36,33,45,36,42,41,39,38,39,38,35,47,51,44,47,40,42,45,47,52,42,53,47,52, 54,50,48,51,46,54,61,56,57,53,49,54,57,60,53,56,53,58,57,57,57,50,52,51,60,61, 62,68,57,62,54,60,59,63,66,70,68,63,68,66,70,63,70,68,66,73,66,67,72,75,70,60, 74,75,66,79,69,75,81,83,67,73,72,74,82,80,70,87,81,80,76,80,71,76, //200 80,82,91,78,71,78,74,82,85,83,89,85,95,82,81,102,78,85,96,85,90,93,85,81,95,87, 102,99,96,89,89,98,93,88,93,94,91,100,96,99,104,100,94,102,103,97,101,100,105, 103,105,104,99,114,110,104,120,120,103,112,116,104,106,121,110,115,109,108,115, 103,117,105,133,128,113,125,108,102,122,115,109,111,109,120,135,126,116,116,117, 124,116,127,120,125,105,122,132,125,123,121, //300 128,124,115,142,131,130,123,133,142,117,136,136,129,134,125,134,129,136,128,133, 138,128,141,142,139,140,155,140,145,154,133,144,137,135,150,137,159,142,155,147, 137,155,144,150,160,144,159,149,147,143,158,147,166,163,153,158,172,162,150,166, 157,161,155,158,166,160,168,156,157,163,157,152,162,151,164,162,166,166,165,166, 167,165,172,154,172,172,172,171,163,172,161,161,170,174,191,169,171,175,179,178,//400 182,193,181,171,167,183,161,180,180,192,185,171,185,178,165,186,177,179,171,194, 179,189,178,189,197,198,184,193,186,197,188,176,202,201,184,199,190,177,205,198, 201,215,204,185,200,211,203,204,186,216,215,200,178,195,196,189,190,193,212,199, 189,202,210,207,199,213,203,191,227,201,215,213,200,199,207,207,225,208,196,198, 235,223,229,225,209,206,202,223,200,244,209,217,233,206,224,227,224,229,223,218,//(500) 221,236,222,222,216,223,221,225,231,233,237,227,220,242,243,217,234,221,236,235, 241,244,228,243,245,238,245,221,227,231,240,221,231,243,234,245,262,248,253,226, 250,244,247,232,237,240,244,233,260,243,270,232,281,264,247,228,248,252,241,264, 255,262,259,255,245,264,259,232,262,271,245,254,252,258,261,266,256,273,273,275, 261,254,270,268,269,255,262,255,273,259,267,286,247,286,268,263,253,249,267,272,//600 278,268,269,270,279,283,298,278,264,266,267,263,265,265,282,268,280,293,293,294, 269,302,294,280,300,295,284,284,304,282,300,292,289,274,303,320,291,286,285,293, 288,296,296,309,288,294,282,319,295,294,294,293,301,304,284,299,299,304,311,300, 311,301,322,297,298,312,322,310,301,330,309,307,334,301,317,314,276,327,338,312, 327,302,336,316,304,307,331,313,308,296,291,294,321,309,317,318,331,304,337,334,//700 327,332,326,313,320,315,338,319,325,314,332,316,331,322,352,315,336,337,333,332, 340,363,321,364,338,348,329,338,347,337,336,341,348,343,368,335,335,370,345,343, 350,319,336,355,347,348,354,346,330,335,336,351,354,366,374,356,363,366,349,348, 366,340,345,346,370,370,366,353,369,363,355,369,368,355,384,352,368,343,379,363, 371,365,368,367,367,382,353,350,378,369,370,370,375,368,361,362,374,388,362,397,//800 365,375,369,385,397,387,385,375,371,383,357,403,379,365,376,383,382,394,406,390, 384,374,393,400,393,382,385,383,391,402,372,386,382,385,389,382,388,419,383,407, 391,400,397,395,396,383,401,390,400,406,402,411,390,410,380,396,383,411,423,403, 392,413,398,378,436,414,417,407,411,415,393,415,421,394,411,419,429,434,431,415, 411,421,407,423,420,410,428,401,450,413,412,439,430,435,431,448,433,415,416,422,//900 443,432,434,442,419,475,405,439,440,442,435,423,418,412,409,431,437,443,436,439, 426,462,421,456,409,440,444,435,441,441,445,424,454,435,455,424,446,454,448,445, 453,446,453,448,437,445,474,445,472,458,445,448,462,450,443,463,468,456,447,460, 464,453,451,481,456,462,467,459,471,461,463,469,451,469,440,460,469,481,456,464, 446,451,473,483,488,492,478,490,457,474,486,467,498,459,496,483,478,466,469,472};int main(){ int n; while(cin>>n) cout<<a[n]<<endl; return 0;}

 POJ 1008解题报告，只是做下笔记。翻译~~~

/***************************************************************************************************************Description上周末，M.A. Ya教授对古老的玛雅有了一个重大发现。从一个古老的节绳（玛雅人用于记事的工具）中，教授发现玛雅人使用了一个一年有365天的叫做Haab的历法。这个Haab历法拥有19个月，在开始的18个月，一个月有20天，月份的名字分别是pop, no, zip, zotz, tzec, xul, yoxkin, mol, chen, yax, zac, ceh, mac, kankin, muan, pax, koyab, cumhu。这些月份中的日期用0到19表示。Haab历的最后一个月叫做uayet，它只有5天，用0到4表示。玛雅人认为这个日期最少的月份是不吉利的，在这个月法庭不开庭，人们不从事交易，甚至没有人打扫屋中的地板。因为宗教的原因，玛雅人还使用了另一个历法，在这个历法中年被称为Tzolkin(holly年)，一年被分成13个不同的时期，每个时期有20天，每一天用一个数字和一个单词相组合的形式来表示。使用的数字是1~13，使用的单词共有20个，它们分别是：imix, ik, akbal, kan, chicchan, cimi, manik, lamat, muluk, ok, chuen, eb, ben, ix, mem, cib, caban, eznab, canac, ahau。注意：年中的每一天都有着明确唯一的描述，比如，在一年的开始，日期如下描述:1 imix,2 ik,3 akbal,4 kan,5 chicchan,6 cimi,7 manik,8 lamat,9 muluk,10 ok,11 chuen,12 eb,13 ben,1 ix,2 mem,3 cib,4 caban,5 eznab,6 canac,7 ahau,8 imix,9 ik,10 akbal ……也就是说数字和单词各自独立循环使用。Haab历和Tzolkin历中的年都用数字0, 1, ……表示，数字0表示世界的开始。所以第一天被表示成：Haab: 0. pop 0Tzolkin: 1 imix 0请帮助M.A. Ya教授写一个程序可以把Haab历转化成Tzolkin历。InputHaab历中的数据由如下的方式表示：日期. 月份 年数输入中的第一行表示要转化的Haab历日期的数据量。下面的每一行表示一个日期，年数小于5000。OutputTzolkin历中的数据由如下的方式表示：天数字 天名称 年数第一行表示输出的日期数量。下面的每一行表示一个输入数据中对应的Tzolkin历中的日期。****************************************************************************************************************/#include<iostream>#include<string>#include<cstring>#include<cstdio>using namespace std;string month,Haab[20]={"pop","no","zip","zotz","tzec","xul","yoxkin","mol","chen","yax",                 "zac","ceh","mac","kankin","muan","pax","koyab","cumhu","uayet"},       Tzokin[20]={"imix","ik","akbal","kan","chicchan","cimi","manik","lamat","muluk",                   "ok","chuen","eb","ben","ix","mem","cib","caban","eznab","canac","ahau"};     int m,num,sum,day,year;                        int main(){   cin>>num;    cout<<num<<endl;    while(num--)    {   scanf("%d.",&day);        cin>>month>>year;        for(m=0;m<19;m++)            if(month.compare(Haab[m])==0) break;        sum=year*365+m*20+day;        year=sum/260;day=(sum-20*(sum-year*260)%20)%13+1;        cout<<day<<" "<<Tzokin[(sum-year*260)%20]<<" "<<year<<endl;    }    return 0;}