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模擬

代碼

#include <cmath>
#include <cstdio>
#include <iostream>
#define N 50
#define M 1000001
#define LL long long

int n, m;
LL f[N], a[M], p[M];
bool b;

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘;
	return x * f;
}

int main()
{
	int i;
	n = read();
	f[1] = 1;
	for(i = 2; i <= n; i++) f[i] = (f[i - 1] + f[i - 2]) % (1 << 31);
	if(f[n] == 1)
	{
		printf("%lld=%lld", f[n], f[n]);
		return 0;
	}
	m = sqrt(f[n]);
	printf("%lld=", f[n]);
	for(i = 2; i <= m; i++)
		while(!(f[n] % i))
		{
			if(!b) printf("%d", i), b = 1;
			else printf("*%d", i);
			f[n] /= i;
		}
	if(b && f[n] > 1) printf("*%lld", f[n]);
	else if(!b && f[n] > 1) printf("%lld", f[n]);
	return 0;
}

[luoguP2626] 斐波那契數列（升級版）（模擬）