rt

本蒟蒻 $TLE$ 4个点，加 $O2$ 会 $WA$ 一个点 $(95)$ 分

以下是代码

#include<bits/stdc++.h>
#define ll long long
const int MOD = 1e9 + 7;
using namespace std;
int a, b, c, d, o, w, len1, len2; string n, m, n1, m1;
struct node{
	ll p[5][5];
	ll* operator[] (int x) {
		return p[x];
	}
} unit, stan, ans1, ans2;
node operator* (node x, node y) {
	node q;
	for(int i = 1; i <= 2; ++i) {
		for(int j = 1; j <= 2; ++j) {
			q[i][j] = 0;
			for(int k = 1; k <= 2; ++k) {
				q[i][j] += 1ll * x[i][k] * y[k][j] % MOD;
				q[i][j] %= MOD;
			}
		}
	} return q;
}
int main() {
	cin >> n >> m; scanf("%d%d%d%d", &a, &b, &c, &d);
	len1 = n.length() - 1, len2 = m.length() - 1;
	o = len2; n1 = n; w = len1; m1 = m;
	ans1[1][1] = 1; ans1[2][2] = 1; stan[1][1] = a;
	stan[1][2] = b, stan[2][1] = 0, stan[2][2] = 1;
	while(m1[o] == '0' && o > -1) m1[o] = '9', --o; m1[o]--;
	while(n1[w] == '0' && w > -1) n1[w] = '9', --o; n1[w]--;
	o = len2; w = len1;
	while(o > -1) {
		node t = stan;
		for(int i = 1; i <= m1[o] - '0'; ++i) ans1 = t * ans1;
		for(int i = 1; i <= 9; ++i) stan = t * stan; --o;
	}
	for(int i = 1; i <= 2; ++i) stan[i][1] = ans1[i][1], stan[i][2] = ans1[i][2];
	ans2[1][1] = 1; ans2[2][2] = 1; (stan[1][1] *= c) %= MOD;
	(((stan[1][2] *= c) %= MOD) += d) %= MOD;
	while(w > -1) {
		node t = stan;
		for(int i = 1; i <= n1[w] - '0'; ++i) ans2 = ans2 * t;
		for(int i = 1; i <= 9; ++i) stan = stan * t; --w;
	}
	node ans; ans[1][1] = 1; ans[2][1] = 1;
	ans = ans1 * ans2 * ans;
	printf("%d\n", ans[1][1] % MOD);
	return 0;
}

望大佬帮助