HDOU-----3579---Hello Kiki扩展欧几里得

最新推荐文章于 2019-10-02 22:22:04 发布

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最新推荐文章于 2019-10-02 22:22:04 发布

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分类专栏：扩展欧几里得

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本文介绍了一个有趣的计数问题——HelloKiki如何通过不同方式计数硬币来找出最少的正整数解。使用欧几里得算法解决此问题，并提供了解决方案的C++代码实现。

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Hello Kiki

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3524 Accepted Submission(s): 1304

Problem Description

One day I was shopping in the supermarket. There was a cashier counting coins seriously when a little kid running and singing "门前大桥下游过一群鸭，快来快来数一数，二四六七八". And then the cashier put the counted coins back morosely and count again...
Hello Kiki is such a lovely girl that she loves doing counting in a different way. For example, when she is counting X coins, she count them N times. Each time she divide the coins into several same sized groups and write down the group size Mi and the number of the remaining coins Ai on her note.
One day Kiki's father found her note and he wanted to know how much coins Kiki was counting.

Input

The first line is T indicating the number of test cases.
Each case contains N on the first line, Mi(1 <= i <= N) on the second line, and corresponding Ai(1 <= i <= N) on the third line.
All numbers in the input and output are integers.
1 <= T <= 100, 1 <= N <= 6, 1 <= Mi <= 50, 0 <= Ai < Mi

Output

For each case output the least positive integer X which Kiki was counting in the sample output format. If there is no solution then output -1.

Sample Input

2
2
14 57
5 56
5
19 54 40 24 80
11 2 36 20 76

Sample Output

Case 1: 341
Case 2: 5996

欧几里得模板，注意最小整数解为0时的情况，因为是存钱，所以不存在为0，所以此时应加上最小公倍数

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<set> 
#include<queue>
#include<vector>
#include<functional>
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CL(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long LL;
const int maxn = 1e5+10;
const int MOD = 1e9+7; 
int n[20], m[20];
void ex_gcd(int a, int b, int& d, int& x, int& y){
	if(!b){
		d = a;
		x = 1;
		y = 0;
	}
	else{
		ex_gcd(b, a%b, d, y, x);
		y -= x*(a/b);
	}
}
int main(){
	int T, d, x, y, m1, m2, m3, n1, n2, n3, v, N, M;
	int kcase = 1;
	scanf("%d", &T);
	while(T--){
		scanf("%d", &N);
		for(int i = 1; i <= N; i++) scanf("%d", &m[i]);
		for(int i = 1; i <= N; i++) scanf("%d", &n[i]);
		bool flag = false;
		v = n1 = n[1]; m1 = m[1];
		for(int i = 2; i <= N; i++){
			m2 = m[i]; n2 = n[i];
			v = max(v, n2);
			ex_gcd(m1, m2, d, x, y);
			LL c, t, k;
			c = n2-n1;
			if(c % d){//注意判断
				flag = true;
				break;
			}
			c /= d;
			t = m2/d;
			x *= c;
			k = (x%t+t)%t;
			n1 += m1*k; 
			m1 = m1/d*m2;
			n1 = (n1%m1+m1)%m1;
		}
		if(n1 < v) n1 += m1;
		printf("Case %d: %d\n", kcase++, flag ? -1 : !n1 ? m1 : n1);//n1为0的情况
	}
	return 0;
}