The number of binary strings that must have 1 on the left of 0
Title description
Given an integer n, find the number of strings with a "1" character on the left side of the "0" character among all character strings consisting of "0" characters and "1" characters in length n.
Enter a description:
Enter a line containing an integer n (1 ≤ n ≤ 2 ∗ 1 0 7) n(1 \leq n \leq2*10^7)n(1≤n≤2∗107)。
Output description:
Output an integer to indicate the returned answer. Due to the huge number of strings, it may overflow. Please output 2 29 2^{29}22 The answer after 9 modulo.
Example 1
enter
1
Output
1
Description
只有“1”满足
Example 2
enter
2
Output
2
Description
只有“10”和“11”满足
Example 3
enter
3
Output
3
Description
只有“101”，“110”，“111”满足
answer:
When we draw on paper, we can find that the first few items are: 1 2 3 5 8 13 …, the obvious Fibonacci sequence. The data range is a bit large, O (n) O(n)O ( n ) should not pass, use matrix fast exponentiation and it's done.
Code:
#include <cstdio>
#include <cstring>
using namespace std;
const long MOD = 1 << 29;
int n;
long c[2][2];
void matmul(long a[][2], long b[][2]) {
memset(c, 0, sizeof c);
for (int i = 0; i < 2; ++i) {
for (int j = 0; j < 2; ++j) {
for (int k = 0; k < 2; ++k) {
c[i][j] += a[i][k] * b[k][j] % MOD;
c[i][j] %= MOD;
}
}
}
memcpy(a, c, sizeof c);
}
int solve(int m) {
long ans[][2] = {
{
1,1}, {
1,0} };
long ret[][2] = {
{
1,0}, {
0,1} };
while (m) {
if (m & 1) matmul(ret, ans);
matmul(ans, ans);
m >>= 1;
}
return ret[0][0];
}
int main(void) {
scanf("%d", &n);
if (n < 3) return 0 * printf("%d\n", n);
printf("%d\n", solve(n));
return 0;
}