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UVa 10515 Powers Et Al.
方法:数论
注意last digit只与m的最后一位和m有关,同时,如果了解number theory 中的multiplicative order 的话,就会发现,1-9 mod 10 的order 不是1 就是4, 所以结果只与m的最后一位和n mod 4的结果有关。而n mod 4 的结果只与 n的最后后两位有关。注意如果n % 4 = 0, 需要n += 4 ,否则会得到m^0 = 1。
code:
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <string> #include <vector> #include <stack> #include <bitset> #include <cstdlib> #include <cmath> #include <set> #include <list> #include <deque> #include <map> #include <queue> #include <fstream> #include <cassert> #include <unordered_map> #include <cmath> #include <sstream> #include <time.h> #include <complex> #include <iomanip> #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) #define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a)) #define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a)) #define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a)) #define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a)) #define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a)) #define FOREACH(a,b) for (auto &(a) : (b)) #define rep(i,n) FOR(i,0,n) #define repn(i,n) FORN(i,1,n) #define drep(i,n) DFOR(i,n-1,0) #define drepn(i,n) DFOR(i,n,1) #define MAX(a,b) a = Max(a,b) #define MIN(a,b) a = Min(a,b) #define SQR(x) ((LL)(x) * (x)) #define Reset(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define mp make_pair #define pb push_back #define all(v) v.begin(),v.end() #define ALLA(arr,sz) arr,arr+sz #define SIZE(v) (int)v.size() #define SORT(v) sort(all(v)) #define REVERSE(v) reverse(ALL(v)) #define SORTA(arr,sz) sort(ALLA(arr,sz)) #define REVERSEA(arr,sz) reverse(ALLA(arr,sz)) #define PERMUTE next_permutation #define TC(t) while(t--) #define forever for(;;) #define PINF 1000000000000 #define newline ‘\n‘ #define test if(1)if(0)cerr using namespace std; using namespace std; typedef vector<int> vi; typedef vector<vi> vvi; typedef pair<int,int> ii; typedef pair<double,double> dd; typedef pair<char,char> cc; typedef vector<ii> vii; typedef long long ll; typedef unsigned long long ull; typedef pair<ll, ll> l4; const double pi = acos(-1.0); string s1, s2; int m, n; int main() { while (cin >> s1 >> s2) { if (s1[0] == ‘0‘ && s2[0] == ‘0‘) break; m = s1.back()-‘0‘; if (m == 0) { cout << 0 << newline; continue; } if (s2[0] == ‘0‘) { cout << 1 << newline; continue; } n = s2.back()-‘0‘; s2.pop_back(); if (s2.length() > 0) n += 10*(s2.back()-‘0‘); n += 4; int ret = m; for (int i = 1; i < n; ++i) ret = ret * m % 10; cout << ret << newline; } }
UVa 10515 Powers Et Al.
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