题目大意:统计位数不大于n的自守数的个数。

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

数据规模:1<=n<=2000。

理论基础:

自守数:参见链接1。

性质定理1:一个数为自守数当且仅当它为一个自守数的后缀。

性质定理2:(1除外)n位数的自守数仅有两个(位数包括前导0),优先考虑最高位不为0的时候。

题目分析:有了两个性质定理以后我们可以知道,每个自守数都是已有的自守数+前缀数字而来的,所以只要知道n位的两个自守数,就可以得到n+1位的两个自守数。反过来思考,如果我们知道了n+1位的自守数后那么小于n+1位的自守数我们都可以写出来。这样就出现了一个想法,先用递归将两个两千位(bign)(仅用到大数的乘法操作不是很困难)的自守数求出来,然后我们再找出值相同的自守数的个数m。用总数:1+2*n减去m即为答案了。当然,打表肯定是最好啦。扫描前导零的个数即为重复的个数。

代码如下:

#include<iostream>
#include<cstdio>
using namespace std;
char number1[2001]=
"0302695456948792438016548848805106486276062082716415913252360\
9790500938385405426324719893931802209823600162545177681029159\
3965045066578090330527721983852863418796455114247485363072354\
5704904450912521423427595549184397398445871252869481982692702\
9255264834903206526851272202961318699947776535481291519857640\
4229681830917734452777232007376038258831727292795636574190144\
4523595431910306357249617898820317578776106213770808096781137\
4931911766563031490205784352509572880668464121069252802275061\
2985116162063840067789794024490238751112586895345495148882006\
7866770234100283954928297028644727362521753544319791185506815\
7264858804852673871684804002188529473022223344541221328464844\
1535937936631336044589403287234784019473575603613462120086753\
7334691331433871735088021260028575298538664393102232655345477\
6845029957025561658143370236502074744856814787872902092412582\
9053012491246688683515876774998917686787157281349408792768945\
2979709777230540335661882819870221063055796723980661119019774\
4642421025136748701117131278125400133690086034889084364023875\
7659368219796261819178335204927041993248752378258671482789053\
4489744014261231703569954841949944461060814620725403655999827\
1588356035049327795540741961849280952093753026852390937562839\
1485716123673519706092242423987770075749557872715597674134589\
9753769551586271888794151630756966881635215504889827170437850\
8028434084412644126821848514157729916034497017892335796684991\
4473895660019325458276780006183298544262328272575561107331606\
9701586498422229125548572987933714786632317240551575610235254\
3994999345608083801190741530060056055744818709692785099775918\
0500754164285277081620113502468060581632761716767652609375280\
5684421448619396049983447280672190667041724009423446619781242\
6690787535944616698508064636137166384049029219341881909581659\
5244778618461409128782984384317032481734288865727376631465191\
0498802944796081467376050395719689371467180137561905546299681\
4764263903953007319108169802938509890062166509580863811000557\
423423230896109004106619977392256259918212890625",number2[2001]=
"9697304543051207561983451151194893513723937917283584086747639\
0209499061614594573675280106068197790176399837454822318970840\
6034954933421909669472278016147136581203544885752514636927645\
4295095549087478576572404450815602601554128747130518017307297\
0744735165096793473148727797038681300052223464518708480142359\
5770318169082265547222767992623961741168272707204363425809855\
5476404568089693642750382101179682421223893786229191903218862\
5068088233436968509794215647490427119331535878930747197724938\
7014883837936159932210205975509761248887413104654504851117993\
2133229765899716045071702971355272637478246455680208814493184\
2735141195147326128315195997811470526977776655458778671535155\
8464062063368663955410596712765215980526424396386537879913246\
2665308668566128264911978739971424701461335606897767344654522\
3154970042974438341856629763497925255143185212127097907587417\
0946987508753311316484123225001082313212842718650591207231054\
7020290222769459664338117180129778936944203276019338880980225\
5357578974863251298882868721874599866309913965110915635976124\
2340631780203738180821664795072958006751247621741328517210946\
5510255985738768296430045158050055538939185379274596344000172\
8411643964950672204459258038150719047906246973147609062437160\
8514283876326480293907757576012229924250442127284402325865410\
0246230448413728111205848369243033118364784495110172829562149\
1971565915587355873178151485842270083965502982107664203315008\
5526104339980674541723219993816701455737671727424438892668393\
0298413501577770874451427012066285213367682759448424389764745\
6005000654391916198809258469939943944255181290307214900224081\
9499245835714722918379886497531939418367238283232347390624719\
4315578551380603950016552719327809332958275990576553380218757\
3309212464055383301491935363862833615950970780658118090418340\
4755221381538590871217015615682967518265711134272623368534808\
9501197055203918532623949604280310628532819862438094453700318\
5235736096046992680891830197061490109937833490419136188999442\
576576769103890995893380022607743740081787109376"; int main()
{
int n,cnt=0;
scanf("%d",&n);
for(int i=1999;i>=1999-(n-1);i--)
{
if(number1[i]=='0')cnt++;
if(number2[i]=='0')cnt++;
}
printf("%d\n",2*n-cnt+1);
return 0;
}

其中number1,number2即为求出的两千位的两个自守数啦。记得打出自守数时一定要左补零。

参考文献:

http://zh.wikipedia.org/wiki/%E8%87%AA%E5%AE%88%E6%95%B0

by:Jsun_moon http://blog.csdn.net/Jsun_moon

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