generator 1(2019年牛客多校第五场B题+十进制矩阵快速幂)
题目链接
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;
}
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