Gym 100548K Last Defence (数论)

题意：给定两个数，然后从第三个开始，每个数都是前两个数的差的绝对值，问这个序列中有多少个不同的元素。

析：这个和辗转相除法差不多，假设a  > b那么a-b之间就有a/b个数，然后再计算a%b-b之间的，直到最后算到0.

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <unordered_map>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10005;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}

int main(){
    int T;  cin >> T;
    LL m, n;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%I64d %I64d", &n, &m);
        printf("Case #%d: ", kase);
        if(!n && !m){ printf("1\n");  continue; }
        if(!n || !m || m == n){ printf("2\n");  continue; }
        LL ans = 1;
        if(m > n) swap(m, n);
        while(m != 0){
            ans += n / m;
            LL tmp = n % m;
            n = m;
            m = tmp;
        }
        printf("%I64d\n", ans);
    }
    return 0;
}