51Nod 1016 水仙花数 V2(组合数学,枚举打表法)
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水仙花数是指一个 n 位数 ( n≥3 ),它的每个位上的数字的 n 次幂之和等于它本身。(例如:1^3 + 5^3 + 3^3 = 153,1634 = 1^4 + 6^4 + 3^4 + 4^4)。
给出一个整数M,求 >= M的最小的水仙花数。
Input
一个整数M(10 <= M <= 10^60)
Output
输出>= M的最小的水仙花数,如果没有符合条件的水仙花数,则输出:No Solution
Input示例
300
Output示例
370
题目链接:http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1016
分析:
一道賊变态的题,如果按常规出牌,绝对做不出来,必然会超时,或者说,我肯定是做不成的。
但是如果投机取巧,那么这就是一道很简单的题,虽然它数据范围高达60位,但是水仙花数却是有限的,只有89个,所以,我们完全可以打表做题。那么剩下的问题就出现了,这些水仙花数是啥?
想知道这些数都是啥,可以选择两种手段,第一暴力解题,获得数据,但是我想这也忒复杂了,虽然是暴力解题,但是依然存在很多问题,就算你写出来代码,想获得所有的水仙花数依然需要很长很长时间等待哦,毕竟运算量惊人。
于是我只好选择第二种办法了,找度娘喽……
这是水仙花数……
1 0 2 1 3 2 4 3 5 4 6 5 7 6 8 7 9 8 10 9 11 153 12 370 13 371 14 407 15 1634 16 8208 17 9474 18 54748 19 92727 20 93084 21 548834 22 1741725 23 4210818 24 9800817 25 9926315 26 24678050 27 24678051 28 88593477 29 146511208 30 472335975 31 534494836 32 912985153 33 4679307774 34 32164049650 35 32164049651 36 40028394225 37 42678290603 38 44708635679 39 49388550606 40 82693916578 41 94204591914 42 28116440335967 43 4338281769391370 44 4338281769391371 45 21897142587612075 46 35641594208964132 47 35875699062250035 48 1517841543307505039 49 3289582984443187032 50 4498128791164624869 51 4929273885928088826 52 63105425988599693916 53 128468643043731391252 54 449177399146038697307 55 21887696841122916288858 56 27879694893054074471405 57 27907865009977052567814 58 28361281321319229463398 59 35452590104031691935943 60 174088005938065293023722 61 188451485447897896036875 62 239313664430041569350093 63 1550475334214501539088894 64 1553242162893771850669378 65 3706907995955475988644380 66 3706907995955475988644381 67 4422095118095899619457938 68 121204998563613372405438066 69 121270696006801314328439376 70 128851796696487777842012787 71 174650464499531377631639254 72 177265453171792792366489765 73 14607640612971980372614873089 74 19008174136254279995012734740 75 19008174136254279995012734741 76 23866716435523975980390369295 77 1145037275765491025924292050346 78 1927890457142960697580636236639 79 2309092682616190307509695338915 80 17333509997782249308725103962772 81 186709961001538790100634132976990 82 186709961001538790100634132976991 83 1122763285329372541592822900204593 84 12639369517103790328947807201478392 85 12679937780272278566303885594196922 86 1219167219625434121569735803609966019 87 12815792078366059955099770545296129367 88 115132219018763992565095597973971522400 89 115132219018763992565095597973971522401
看着好爽啊,这么长……
接下来要考虑的问题就是比大小,这也好解决,位数不同的比位数,相同的再逐位比大小。大概就是这个样子吧。剩下的就没有什么难点了。
无耻的打表徒,来一发AC:
1 #include <stdio.h> 2 #include <string.h> 3 4 int compare(char *a, char *b, int len) 5 { 6 for (int i = 0; i < len; i++) 7 { 8 if (b[i] > a[i]) 9 { 10 return 1; 11 } 12 else if (b[i] < a[i]) 13 { 14 return 0; 15 } 16 } 17 return 1; 18 } 19 20 int main(int argc, const char * argv[]) 21 { 22 char NarNum[89][60] = {"0","1","2","3","4","5","6","7","8","9","153","370","371","407","1634","8208","9474","54748","92727","93084","548834","1741725","4210818","9800817","9926315","24678050","24678051","88593477","146511208","472335975","534494836","912985153","4679307774","32164049650","32164049651","40028394225","42678290603","44708635679","49388550606","82693916578","94204591914","28116440335967","4338281769391370","4338281769391371","21897142587612075","35641594208964132","35875699062250035","1517841543307505039","3289582984443187032","4498128791164624869","4929273885928088826","63105425988599693916","128468643043731391252","449177399146038697307","21887696841122916288858","27879694893054074471405","27907865009977052567814","28361281321319229463398","35452590104031691935943","174088005938065293023722","188451485447897896036875","239313664430041569350093","1550475334214501539088894","1553242162893771850669378","3706907995955475988644380","3706907995955475988644381","4422095118095899619457938","121204998563613372405438066","121270696006801314328439376","128851796696487777842012787","174650464499531377631639254","177265453171792792366489765","14607640612971980372614873089","19008174136254279995012734740","19008174136254279995012734741","23866716435523975980390369295","1145037275765491025924292050346","1927890457142960697580636236639","2309092682616190307509695338915","17333509997782249308725103962772","186709961001538790100634132976990","186709961001538790100634132976991","1122763285329372541592822900204593","12639369517103790328947807201478392","12679937780272278566303885594196922","1219167219625434121569735803609966019","12815792078366059955099770545296129367","115132219018763992565095597973971522400","115132219018763992565095597973971522401"}; 23 int NarNumLen; 24 char num[60]; 25 scanf("%s", num); 26 int NumLen = (int)strlen(num); 27 for (int i = 0; i < 89; i++) 28 { 29 NarNumLen = (int)strlen(NarNum[i]); 30 if ((NumLen == NarNumLen && compare(num, NarNum[i], NumLen)) || NumLen < NarNumLen) 31 { 32 printf("%s\n", NarNum[i]); 33 return 0; 34 } 35 } 36 37 puts("No Solution"); 38 return 0; 39 }
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