Codeforces 678D Iterated Linear Function

链接

传送门

题意

定义$f(x)=Ax+B$、$g^{(0)}(x)=x$、$g^{(n)}(x)=g^{(n-1)}(x)$。给出$A$、$B$、$n$、$x$ $(1 \leq A,B,x \leq 10^9,1 \leq n \leq 10^{18})$,求$g^{(n)}(x)$对$10^9+7$取模的值。

思路

数据范围和题目形式很显然的就是矩阵题。

代码

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#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL mod = 1000000007;
const int maxn = 2;
struct Matrix {
LL a[maxn][maxn];
Matrix() {
memset(a, 0, sizeof a);
for (int i = 0; i < maxn; ++i) {
a[i][i] = 1;
}
}
Matrix operator * (const Matrix& rhs) const {
Matrix res;
memset(res.a, 0, sizeof res.a);
for (int k = 0; k < maxn; ++k) {
for (int i = 0; i < maxn; ++i) {
for (int j = 0; j < maxn; ++j) {
res.a[i][j] = (res.a[i][j] + a[i][k] * rhs.a[k][j]) % mod;
}
}
}
return res;
}
};
Matrix pow_mod(Matrix x, LL k) {
Matrix res, cur = x;
while (k) {
if (k & 1) {
res = res * cur;
}
cur = cur * cur;
k >>= 1;
}
return res;
}
int main() {
LL A, B, n, x;
scanf("%lld%lld%lld%lld", &A, &B, &n, &x);
Matrix ans;
ans.a[0][0] = A, ans.a[0][1] = B;
ans.a[1][0] = 0, ans.a[1][1] = 1;
ans = pow_mod(ans, n);
printf("%lld\n", (ans.a[0][0] * x + ans.a[0][1]) % mod);
return 0;
}

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