HDU1848 Fibonacci again and again

解：先搞出斐波那契数列，然后n²暴力算一堆时的SG函数，然后异或起来回答询问。

#include <cstdio>
#include <cstring>

const int N = 1010;

int sg[N], f[N], bin[N];

int main() {
    f[1] = 1;
    f[2] = 2;
    for(int i = 3; f[i - 1] <= 1000; i++) {
        f[i] = f[i - 1] + f[i - 2];
    }
    sg[0] = 0;
    for(int i = 1; i <= 1000; i++) {
        if(i > 1) memset(bin, 0, sizeof(bin));
        for(int j = 1; f[j] <= i; j++) {
            bin[sg[i - f[j]]]++;
            //printf("bin %d ++ 
", sg[i - f[j]]);
        }
        for(int j = 0; ; j++) {
            //printf("bin %d = %d 
", j, bin[j]);
            if(!bin[j]) {
                sg[i] = j;
                break;
            }
        }
        //printf("sg %d = %d 
", i, sg[i]);
    }

    int a, b, c;
    while(scanf("%d%d%d", &a, &b, &c)) {
        if(!(a | b | c)) break;
        int t = sg[a] ^ sg[b] ^ sg[c];
        if(!t) printf("Nacci
");
        else printf("Fibo
");
    }

    return 0;
}