generator 1(2019年牛客多校第五场B题+十进制矩阵快速幂)

目录

• 题目链接
• 思路
• 代码

题目链接

传送门

思路

十进制矩阵快速幂。

代码

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson (rt<<1),L,mid
#define rson (rt<<1|1),mid + 1,R
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("/home/dillonh/CLionProjects/Dillonh/in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 1000000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

int MOD;
char s[maxn];

struct matrix {
    int a[2][2];
    matrix operator * (const matrix& x) const {
        matrix b;
        for(int i = 0; i < 2; ++i) {
            for(int j = 0; j < 2; ++j) {
                b.a[i][j] = 0;
                for(int k = 0; k < 2; ++k) {
                    b.a[i][j] = (b.a[i][j] + 1LL * a[i][k] * x.a[k][j] % MOD) % MOD;
                }
            }
        }
        return b;
    }
}x, A[10];

matrix qpow(matrix x, int n) {
    matrix b;
    memset(b.a, 0, sizeof(b.a));
    b.a[0][0] = b.a[1][1] = 1;
    while(n) {
        if(n & 1) b = b * x;
        x = x * x;
        n >>= 1;
    }
    return b;
}

int main() {
#ifndef ONLINE_JUDGE
    FIN;
#endif
    memset(x.a, 0, sizeof(x.a));
    scanf("%d%d%d%d", &x.a[0][1], &x.a[0][0], &A[1].a[0][0], &A[1].a[1][0]);
    A[1].a[0][1] = 1;
    scanf("%s%d", s + 1, &MOD);
    for(int i = 2; i <= 9; ++i) A[i] = A[i-1] * A[1];
    int n = strlen(s + 1);
    matrix ans = A[s[1]-'0'];
    for(int i = 2; i <= n; ++i) {
        ans = qpow(ans, 10);
        if(s[i] > '0') {
            ans = ans * A[s[i]-'0'];
        }
    }
    ans = x * ans;
    printf("%d\n", ans.a[0][1]);
    return 0;
}