HDU1097

ACM-数论之A hard puzzle——hdu1097

AhardpuzzleProblemDescriptionlcygivesahardpuzzletofeng5166,lwg,JGShiningandIgnatius:gaveaandb,howtoknowthea^b.everybodyobjectstothisBTproblem,solcymakestheproblemeasierthanbegin.thispuzzledescribestha
ltree98·2020-09-15 14:59

HDU1097 A hard puzzle

点这里题意：算a的b次方的最后一位数。#includeusingnamespacestd;intmain(){inta,b;while(~scanf("%d%d",&a,&b)){a%=10;b%=4;if(!b)b=4;a=pow(a,b);printf("%d\n",a%10);}return0;}
伊莎贝拉•狗剩·2020-09-14 04:59

HDU1097

Descriptionlcygivesahardpuzzletofeng5166,lwg,JGShiningandIgnatius:gaveaandb,howtoknowthea^b.everybodyobjectstothisBTproblem,solcymakestheproblemeasierthanbegin. thispuzzledescribesthat:gaveaandb,howto
longshanxiaoxuesheng·2015-11-29 12:00

hdu1097求幂的个位

想通了的话这道题目很easy。其实对这种什么取个位，取模一个很小的数的都应当想到打表。像这个题目就完全可以作为一个模版使用了。。 #include<iostream> using namespace std; const int tab[10][4]={{0,0,0,0},{1,1,1,1},{2,4,8,6},{3,9,7,1},{4,6,4,6},{5,5,5,5},{
·2015-11-08 10:35

Hdu1097（计算a的b次幂最后一位数值）

  1 #include <stdio.h> 2 #include <math.h> 3 int main() 4 { 5 int Num1,Num2; 6 while(scanf("%d %d",&Num1,&Num2)!=EOF) 7 { 8 p
·2015-11-03 22:33

简单的快速幂取模

 HDU1097本题为快速幂取模运算，直接套用模板即可//题目描述：给你数a,b,让你输出a^b的最后一位#include #include usingnamespacestd; #defineLLlonglong
xiaotan1314·2015-08-26 16:00

hdu1097 A hard puzzle

AhardpuzzleTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):27417    AcceptedSubmission(s):9761ProblemDescriptionlcygivesahardpuzzletofeng5166,lwg,JGShinin
svtter·2014-10-17 23:00

hdu1097 A hard puzzle

AhardpuzzleTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):27417    AcceptedSubmission(s):9761ProblemDescriptionlcygivesahardpuzzletofeng5166,lwg,JGShinin
svtter·2014-10-17 23:00

HDU1097 A hard puzzle

AhardpuzzleTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):29415    AcceptedSubmission(s):10581ProblemDescriptionlcygivesahardpuzzletofeng5166,lwg,JGShini
u012846486·2014-08-27 21:00

HDU1097 A hard puzzle

#includeusingnamespacestd;intquick_mod(inta,intb){  intans=1;  while(b>=1)  {    if(b&1)     ans=((ans%10)*(a%10))%10;//如果不分开取模（ans=ans*a%10）就有可能越界，提交就会WA；    b=b>>1;    a=((a%10)*(a%10))%10;//同ans的取模
AC_Gibson·2014-08-11 12:00

hdu1097

ProblemDescriptionlcygivesahardpuzzletofeng5166,lwg,JGShiningandIgnatius:gaveaandb,howtoknowthea^b.everybodyobjectstothisBTproblem,solcymakestheproblemeasierthanbegin.thispuzzledescribesthat:gaveaandb
u011459262·2013-07-20 11:00

hdu1097 A hard puzzle

#include #include usingnamespacestd; longf(longa,longb,longm); intmain() { longa,b; while(cin>>a>>b) { cout<
wuli2496·2012-05-20 00:00

HDU1097

http://acm.hdu.edu.cn/showproblem.php?pid=1097一道找规律的数学题，a^b的最后一位数，其实这个数是有规律的，在1到9中，最长的周期是4，就是说，4^1=4,4^2=6,4^3=4,4^4=6，结构都是写最后以为而已，4的周期是3，最长的是4，所以就以4为周期来求就可以了，注意要用long的类型，不然数据就爆了#include usingnamespac
Kay_Sprint·2011-10-20 19:00

A hard puzzle ACM HDU1097

A hard puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13437    Accepted Submission(s): 4677 Pr
·2011-08-01 10:00

HDU 1097

HDU1097 AhardpuzzleTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission
tctony·2007-11-29 12:00
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