标签:tor scanf for ges pac mit ret .com lib
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 175 Accepted Submission(s): 119
暴力发现当4 12 33 88 232
和斐波那契数列对比 答案为 第2*k+3个数减1
直接用矩阵快速幂求的F[2*k+3] 然后减1
A=1,B=0;
然后矩阵快速幂2*k+3-1次得到F[2*k+3]
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<string.h> 7 #include<map> 8 #include<set> 9 #include<queue> 10 #include<vector> 11 #include<cstdlib> 12 typedef long long ll; 13 typedef unsigned long long LL; 14 using namespace std; 15 const int INF=0x3f3f3f3f; 16 const int num=100; 17 const int mod=998244353; 18 int N; 19 struct Mat{ 20 ll a[num][num]; 21 void init(){ 22 memset(a,0,sizeof(a)); 23 for(int i=0;i<num;i++) 24 a[i][i]=1; 25 } 26 }; 27 Mat mul(Mat a,Mat b){ 28 Mat ans; 29 for(int i=0;i<N;i++){ 30 for(int j=0;j<N;j++){ 31 ans.a[i][j]=0; 32 for(int k=0;k<N;k++){ 33 ans.a[i][j]+=a.a[i][k]*b.a[k][j]; 34 } 35 ans.a[i][j]=ans.a[i][j]%mod; 36 } 37 } 38 return ans; 39 } 40 Mat power(Mat a,int n){ 41 Mat ans; 42 ans.init(); 43 while(n){ 44 if(n&1){ 45 ans=mul(ans,a); 46 } 47 n=n>>1; 48 a=mul(a,a); 49 } 50 return ans; 51 } 52 int main(){ 53 int k; 54 N=2; 55 while(scanf("%d",&k)!=EOF){ 56 k=k*2+3; 57 Mat aa; 58 aa.a[0][0]=1; 59 aa.a[0][1]=1; 60 aa.a[1][0]=1; 61 aa.a[1][1]=0; 62 Mat ans=power(aa,k-1); 63 ll t=((ans.a[0][0]-1)%mod+mod)%mod; 64 cout<<t<<endl; 65 } 66 return 0; 67 }
标签:tor scanf for ges pac mit ret .com lib
原文地址:http://www.cnblogs.com/Aa1039510121/p/7502798.html