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[luoguP2626] 斐波那契数列(升级版)(模拟)

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代码

#include <cmath>#include <cstdio>#include <iostream>#define N 50#define M 1000001#define LL long longint 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] 斐波那契数列(升级版)(模拟)