https://ac.nowcoder.com/acm/contest/331/J

C++版本一

題解：

考慮每一位，只有在(0,0)(0,1)(1,0)的三種情況時滿足條件。

根據乘法原理，答案即為3^M

#include <bits/stdc++.h> using namespace std; typedef long long ll;const int mod = 998244353;ll n;ll ans[105];int main() {scanf("%lld", &n);ans[0] = 1;for (int i = 1; i <= 100; i++) ans[i] = ans[i - 1] * 3 % mod;printf("%lld\n", ans[n]);return 0; }

C++版本二

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int MOD=998244353; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,q; int ans,cnt,flag,temp; int a[N]; char str; ll PowerMod(ll a, ll b, ll c){ll ans = 1;a = a % c;while(b>0){if(b % 2 == 1)ans = (ans * a) % c;b >>= 1;a = (a * a) % c;}return ans; }int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endifscanf("%d",&n);cout << PowerMod(3,n,MOD) << endl;//cout << "Hello world!" << endl;return 0; }

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