中國剩餘定理+擴展歐幾里得求逆元 hdu 1006

中國剩餘定理

擴展歐幾里得求逆元：

int exgcd(int a,int b,int &x,int &y){
	if(b==0){
		x = 1;
		y = 0;
		return a;
	}
	int r = exgcd(b,a%b,x,y);
	int tmp = x;
	x = y;
	y = tmp - a/b*y;
	return x;
}

求得的逆元可能爲負數記得取模

下面給出代碼

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <climits>
#include <string>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <sstream>
#include <cctype>
using namespace std;
typedef long long ll;
typedef pair<int ,int> pii;
#define MEM(a,b) memset(a,b,sizeof a)
#define CLR(a) memset(a,0,sizeof a);
int exgcd(int a,int b,int &x,int &y){
	if(b==0){
		x = 1;
		y = 0;
		return a;
	}
	int r = exgcd(b,a%b,x,y);
	int tmp = x;
	x = y;
	y = tmp - a/b*y;
	return x;
}
int main()
{
	int kase = 1, p,e,i,d;
	while(cin >>  p >> e >> i >> d && (p!=-1 || e!=-1 || i!=-1 || d!=-1)){
		int x,y;
		int sum = 0;
		sum += (28*33*exgcd(33*28,23,x,y)%(23*28*33)+23*28*33)*p;
		sum += (23*33*exgcd(23*33,28,x,y)%(23*28*33)+23*28*33)*e;
		sum += (23*28*exgcd(23*28,33,x,y)%(23*28*33)+23*28*33)*i;
		sum %= 23*28*33;
		sum = (sum - d + 23*28*33*2)%(23*28*33);
		if(sum){
			printf("Case %d: the next triple peak occurs in %d days.\n",kase++,sum);
		}
		else{
			printf("Case %d: the next triple peak occurs in %d days.\n",kase++,23*33*28);
		}

	}
	return 0;
}