10.25 斐波那契

思路：

对于两个点，从fib数列的后面开始减，一直减的相等

//std 80， 一堆bug

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>

using namespace std;
const int N = 95;

#define LL long long

LL f[N];
LL m;

inline LL read()
{
    LL x = 0; char c = getchar();
    while(c < '0' || c > '9') c = getchar();
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x;
}

int main()
{
    freopen("fibonacci.in", "r", stdin);
    freopen("fibonacci.out", "w", stdout);
    f[1] = 1;
    f[2] = 1;
    for(int i = 3; i <= 90; i ++)
        f[i] = f[i - 1] + f[i - 2];
    m = read();
    for(int i = 1; i <= m; i ++)
    {
        LL a = read();
        LL b = read();
        if(a < b) swap(a, b);//baozheng a > b;
        bool flag = 1;
        int js = 90;
        while(a != b)
        {
            if(a > f[js]) a -= f[js];
            if(b > f[js]) b -= f[js];
            js --;
        }
        printf("%I64d
", a);
    } 
    
    return 0;
}
/*
5
1 1
2 3
5 7
7 13
4 12
*/