POJ - 2115 - C Looooops （扩展欧几里得）

标签：acm   poj   扩展欧几里得   

Description

for (variable = A; variable != B; variable += C)

  statement;

Input

Output

Sample Input

3 3 2 16
3 7 2 16
7 3 2 16
3 4 2 16
0 0 0 0

Sample Output

0
2
32766
FOREVER

Source

还是扩展欧几里得，这里注意要简化一下原来的式子

AC代码：

#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <cctype>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define INF 0x7fffffff
using namespace std;

LL gcd(LL a, LL b) {
	return b == 0 ? a : gcd(b, a % b);
}

void exgcd(LL a, LL b, LL& x, LL& y) {
	if(b == 0) {
		x = 1; y = 0;
	}
	else {
		exgcd(b, a % b, y, x);
		y -= x * (a / b);
	}
}


int main() {
	LL A, B, C, k;
	while(scanf("%I64d %I64d %I64d %I64d", &A, &B, &C, &k) != EOF) {
		if(A == 0 && B == 0 && C == 0 && k == 0) break;
		
		if(A == B) {
			printf("0\n");
			continue;
		}
		
		LL a = C;
		LL b = (1LL << k);
		LL c = gcd(a, b);
		LL d = B - A;
		
		if(d % c != 0) {
			printf("FOREVER\n");
			continue;
		}
		
		a /= c;//这里要进行简化，因为可能产生多余的次数 
		b /= c;
		d /= c;
		
		LL p, q;
		exgcd(a, b, p, q);//这里求的是最简ax+by=gcd(a,b)的一组x,y的解 
		
		printf("%I64d\n", (p * (d / gcd(a, b)) % b + b) % b);
		
	}
	return 0;
}

POJ - 2115 - C Looooops （扩展欧几里得）

标签：acm   poj   扩展欧几里得   

原文地址：http://blog.csdn.net/u014355480/article/details/47084159