題意：矩陣快速冪求斐波那契數列。

分析：

#include<cstdio> #include<cstring> #include<cctype> #include<cstdlib> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<deque> #include<queue> #include<stack> #include<list> #define fin freopen("in.txt", "r", stdin) #define fout freopen("out.txt", "w", stdout) #define pr(x) cout << #x << " : " << x << " " #define prln(x) cout << #x << " : " << x << endl typedef long long ll; typedef unsigned long long llu; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const ll LL_INF = 0x3f3f3f3f3f3f3f3f; const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f; const double pi = acos(-1.0); const double EPS = 1e-6; const int dx[] = {0, 0, -1, 1}; const int dy[] = {-1, 1, 0, 0}; const ll MOD = 1e9; const int MAXN = 1000 + 10; const int MAXT = 10000 + 10; using namespace std; struct Node {ll c[2][2];Node(){memset(c, 0, sizeof c);} }; Node multi(Node &a, Node &b) {Node ans;for(int i = 0; i < 2; ++i)for(int j = 0; j < 2; ++j)for(int k = 0; k < 2; ++k){ans.c[i][j] += (a.c[i][k] * b.c[k][j]);ans.c[i][j] %= MOD;}return ans; } ll q_pow(ll cur) {Node a;a.c[0][0] = a.c[0][1] = a.c[1][0] = 1;Node ans;ans.c[0][0] = ans.c[1][1] = 1;while(cur){if(cur & 1){ans = multi(ans, a);}a = multi(a, a);cur /= 2;}return ans.c[0][0]; } int main() {int P;scanf("%d", &P);while(P--){int K;ll Y;scanf("%d%lld", &K, &Y);printf("%d %lld\n", K, q_pow(Y - 1));}return 0; }

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轉載于:https://www.cnblogs.com/tyty-Somnuspoppy/p/5971287.html

總結

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