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a[i] = (1-a[i-1])/2递推式推出正式公式 然后快速幂加逆元求出公式结果就行了#include<cstdio>#include<algorithm>#include<iostream>#include<cstring>#define mod 1000000007ll#define LL long longusing namespace std;LL n;LL fast

用线段树储存价值和，因为矩阵 a1 * b + a2 * b + a3 * b + a4 * b + a5 * b = (a1 + a2 + a3 + a4 + a5) * b; 然后进行下优化，尽量减少快速幂的运算数量#include<cstdio>#include<algorithm>#include<iostream>#include<cstring>#define lson i<