题解 | 斐波那契数列

#include <stdio.h>

int main() {
    int n;
    scanf("%d", &n);
    int fib[n + 1];
    fib[0] = 0;
    fib[1] = 1, fib[2] = 1;
    if (n == 1 || n == 2) {
        printf("1");
        return 0;
    } else {
        for (int j = 2; j <= n; j++) {
            fib[j] = fib[j - 1] + fib[j - 2];
        }
    }
    printf("%d",fib[n]);
    return 0;
}