POJ 2115 C Looooops

标签：poj   扩展欧几里德算法   

题目：http://poj.org/problem?id=2115

题意：对于C的for(i=A ; i!=B ;i +=C)循环语句，问在k位存储系统中循环几次才会结束。若在有限次内结束，则输出循环次数。否则输出死循环。

思路：这道题是一个扩展欧几里德算法的拓展，求单变元模线性方程 即：Cx=(B-A)(mod 2^k) 

扩展欧几里得算法和单变元模线性方程（传送门） + 比较详细的博客

代码：

/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, n) for (int i = 0; i < n; i++)
#define debug puts("===============")
#define eps (1e-6)
typedef long long ll;
using namespace std;
ll extend_gcd(ll a, ll b, ll &x, ll &y) {
    if (b == 0) {
        x = 1, y = 0;
        return a;
    }
    else {
        ll r = extend_gcd(b, a % b, y, x);
        y -= x * (a / b);
        return r;
    }
}
vector<ll> line_mod_equation(ll a, ll b, ll n) {
    ll x, y;
    ll d = extend_gcd(a, n, x, y);
    vector<ll> ans;
    ans.clear();
    if (b % d == 0) {
        x = (x % n + n) % n;
        x %= (n / d);
        ans.push_back(x * (b / d) % (n / d));
        //for (ll i = 1; i < d; i++) ans.push_back((ans[0] + i * n / d) % n);
    }
    return ans;
}
int main () {
    //freopen("2.txt", "w", stdout);
    ll a, b, c, k;
    while(scanf("%lld%lld%lld%lld", &a, &b, &c, &k) , a || b || c || k) {
        ll n = 1LL << k;
        ll p = ((b - a) % n + n) % n;
        vector<ll> ans = line_mod_equation(c, p, n);
        if (ans.size() == 0) puts("FOREVER");
        else printf("%lld\n", ans[0]);
    }
    return 0;
}

POJ 2115 C Looooops

标签：poj   扩展欧几里德算法   

原文地址：http://blog.csdn.net/sio__five/article/details/38150957