题解 | 斐波那契数列
斐波那契数列
https://www.nowcoder.com/practice/ee5d403c1172487f8c7915b3c3d924c6
#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;
}
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