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Fibonacci again and againTime Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11745 Accepted Submission(s): 5074 Problem Description 任何一个大学生对菲波那契数列(Fibonacci numbers)应该都不会陌生,它是这样定义的: Input 输入数据包含多个测试用例,每个测试用例占一行,包含3个整数m,n,p(1<=m,n,p<=1000)。 Output 如果先手的人能赢,请输出“Fibo”,否则请输出“Nacci”,每个实例的输出占一行。 Sample Input Sample Output Author lcy Source ACM Short Term Exam_2007/12/13 Recommend lcy | We have carefully selected several similar problems for you: 1850 1849 1846 2149 2188 |
正常的sg函数打表即可,结果异或三堆石子的值判断
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define clr(a) memset(a,0,sizeof(a))
#define line cout<<"-------------"<<endl
typedef long long ll;
const int maxn = 1e5+10;
const int MAXN = 1e6+10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9+7;
const int N = 1010;
int n, m, p, num;
int f[N], sg[N], vis[N];
void get_f(){
num = 20;
f[0] = 1; f[1] = 1;
for(int i=2; i<=16; i++)
f[i] = f[i-1] + f[i-2];
}
void get_sg(){
clr(sg);
for(int i=1; i<N; i++){
clr(vis);
for(int j=1; f[j]<=i && j<=num; j++)
vis[sg[i-f[j]]] = 1;
for(int j=0; ; j++){
if(!vis[j]){
sg[i] = j; break;
}
}
}
}
int main(){
get_f();
get_sg();
while(scanf("%d%d%d", &n, &m, &p), (n||m||p)){
if(sg[n]^sg[m]^sg[p]) puts("Fibo");
else puts("Nacci");
}
return 0;
}