hdu 4002 欧拉函数 2011大连赛区网络赛B
2015-07-27 17:14
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题意:求1-n内最大的x/phi(x)
通式:φ(x)=x*(1-1/p1)*(1-1/p2)*(1-1/p3)*(1-1/p4)…..(1-1/pn),其中p1, p2……pn为x的所有质因数,x是不为0的整数。φ(1)=1(唯一和1互质的数就是1本身)。
因此含质因数最多的即为所求,打表求出前n个积,之后找到比自己小的最大积
大数打表
2015-07-27:到此一游
通式:φ(x)=x*(1-1/p1)*(1-1/p2)*(1-1/p3)*(1-1/p4)…..(1-1/pn),其中p1, p2……pn为x的所有质因数,x是不为0的整数。φ(1)=1(唯一和1互质的数就是1本身)。
因此含质因数最多的即为所求,打表求出前n个积,之后找到比自己小的最大积
大数打表
2015-07-27:到此一游
#include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<queue> using namespace std; int n,m,t; char c[][500]= { "2", "6", "30", "210", "2310", "30030", "510510", "9699690", "223092870", "6469693230", "200560490130", "7420738134810", "304250263527210", "13082761331670030", "614889782588491410", "32589158477190044730", "1922760350154212639070", "117288381359406970983270", "7858321551080267055879090", "557940830126698960967415390", "40729680599249024150621323470", "3217644767340672907899084554130", "267064515689275851355624017992790", "23768741896345550770650537601358310", "2305567963945518424753102147331756070", "232862364358497360900063316880507363070", "23984823528925228172706521638692258396210", "2566376117594999414479597815340071648394470", "279734996817854936178276161872067809674997230", "31610054640417607788145206291543662493274686990", "4014476939333036189094441199026045136645885247730", "525896479052627740771371797072411912900610967452630", "72047817630210000485677936198920432067383702541010310", "10014646650599190067509233131649940057366334653200433090", "1492182350939279320058875736615841068547583863326864530410", "225319534991831177328890236228992001350685163362356544091910", "35375166993717494840635767087951744212057570647889977422429870", "5766152219975951659023630035336134306565384015606066319856068810", "962947420735983927056946215901134429196419130606213075415963491270", "166589903787325219380851695350896256250980509594874862046961683989710", "29819592777931214269172453467810429868925511217482600306406141434158090", "5397346292805549782720214077673687806275517530364350655459511599582614290", "1030893141925860008499560888835674370998623848299590975192766715520279329390", "198962376391690981640415251545285153602734402721821058212203976095413910572270", "39195588149163123383161804554421175259738677336198748467804183290796540382737190", "7799922041683461553249199106329813876687996789903550945093032474868511536164700810", "1645783550795210387735581011435590727981167322669649249414629852197255934130751870910", "367009731827331916465034565550136732339800312955331782619462457039988073311157667212930", "83311209124804345037562846379881038241134671040860314654617977748077292641632790457335110", "19078266889580195013601891820992757757219839668357012055907516904309700014933909014729740190", "4445236185272185438169240794291312557432222642727183809026451438704160103479600800432029464270", "1062411448280052319722448549835623701226301211611796930357321893850294264731624591303255041960530", "256041159035492609053110100510385311995538591998443060216114576417920917800321526504084465112487730", "64266330917908644872330635228106713310880186591609208114244758680898150367880703152525200743234420230" }; int main() { int i,j,k; //freopen("1.in","r",stdin); scanf("%d",&t); char s[110]; while(t--) { scanf("%s",s); int len1=strlen(s); for(i=1;i<60;i++) { int len2=strlen(c[i]); if(len1>len2) continue; if(len2>len1) break; if(strcmp(s,c[i])<0) break; //比cc里的数小 } printf("%s\n",c[i-1]); } return 0; }
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