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| Time Limit: 1000MS | Memory Limit: 32768KB | 64bit IO Format: %I64d & %I64u |
Description
Input
Output
Sample Input
7 66 8 800
Sample Output
9 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 | #include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cctype>#include <string>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>using namespace std;typedef long long LL;#define CLR(x,y) memset((x),(y),sizeof((x)))#define getint(x) int (x);scanf("%d",&(x))#define get2int(x,y) int (x),(y);scanf("%d%d",&(x),&(y))#define get3int(x,y,z) int (x),(y),(z);scanf("%d%d%d",&(x),&(y),&(z))#define getll(x) LL (x);scanf("%I64d",&(x))#define get2ll(x,y) LL (x),(y);scanf("%I64d%I64d",&(x),&(y))#define get3ll(x,y,z) LL (x),(y),(z);scanf("%I64d%I64d%I64d",&(x),&(y),&(z))#define getdb(x) double (x);scanf("%lf",&(x))#define get2db(x,y) double (x),(y);scanf("%lf%lf",&(x),&(y))#define get3db(x,y,z) double (x),(y),(z);scanf("%lf%lf%lf",&(x),&(y),&(z))#define getint2(x) scanf("%d",&(x))#define get2int2(x,y) scanf("%d%d",&(x),&(y))#define get3int2(x,y,z) scanf("%d%d%d",&(x),&(y),&(z))#define getll2(x) scanf("%I64d",&(x))#define get2ll2(x,y) scanf("%I64d%I64d",&(x),&(y))#define get3ll2(x,y,z) scanf("%I64d%I64d%I64d",&(x),&(y),&(z))#define getdb2(x) scanf("%lf",&(x))#define get2db2(x,y) scanf("%lf%lf",&(x),&(y))#define get3db2(x,y,z) scanf("%lf%lf%lf",&(x),&(y),&(z))#define getstr(str) scanf("%s",str)#define get2str(str1,str2) scanf("%s",str1,str2)#define FOR(x,y,z) for(int (x)=(y);(x)<(z);(x)++)#define FORD(x,y,z) for(int (x)=(y);(x)>=(z);(x)--)#define FOR2(x,y,z) for((x)=(y);(x)<(z);(x)++)#define FORD2(x,y,z) for((x)=(y);(x)>=(z);(x)--)const int maxn = 100;int res[maxn];int main(){ int a,b; while(~get2int2(a,b)) { if(b == 0) { printf("1\n"); continue; } a = a % 10; int t; res[0] = a; for(t = 1;t < maxn;t++) { res[t] = (res[t-1] * a)%10; if(res[t] == a) break; } printf("%d\n",res[ (b-1) % t ]); } return 0;} |
[2016-02-05][HDU][1097][A hard puzzle]
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原文地址：http://www.cnblogs.com/qhy285571052/p/5182563.html
