2017 ACM/ICPC Asia Regional Shenyang Online//number number number
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题目链接:number number number
思路:一个个推,发现K=1时,N=F5-1=4;K=2,N=F7-1=12;K=3,N-F9-1=33.推断出N=F(2*K+3)-1,就是求2*K+3的斐波那契数列。不过这里是要用到矩阵乘法求斐波那契数列。用一般的递归都是会超时的,这里再贴一题也是关于类似的题目。可以把这个直接当成模板题。poj3070
贴代码先吧,分解出来内容太多了,可以直接搜矩阵乘法学习下。
#include <iostream>#include <cstring>using namespace std;const int MOD = 998244353;long long n,A[2][2],res[2][2],c[2][2]; void Mull(long long a[2][2],long long b[2][2]){ memset(c, 0, sizeof(c)); for (int i = 0; i < 2; i++) { for (int j = 0; j < 2; j++) { for (int k = 0; k < 2; k++) { c[i][j] = (c[i][j] + a[i][k] * b[k][j] % MOD) % MOD; } } } memcpy(a, c, sizeof(c));}void Mull_Pow(long long n){ res[0][0] = res[1][1] = 1; res[1][0] = res[0][1] = 0; while (n) { if (n & 1) { Mull(res,A); } Mull(A, A); n >>= 1; }}int main(){ while (~scanf("%lld",&n)) { n=(n-1)*2+4; A[0][0]=1,A[0][1]=1; A[1][0]=1,A[1][1]=0; Mull_Pow(n); printf("%lld\n",(res[0][0]-1+MOD)%MOD);//取模 } return 0;}
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