Number Transformation II（Codeforces Round #201 (Div. 2)）

最新推荐文章于 2020-11-25 19:02:34 发布

Shmilky

最新推荐文章于 2020-11-25 19:02:34 发布

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文章标签： Number Transformation II

原文链接：https://blog.youkuaiyun.com/Fun_Zero/article/details/51170748

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本文解析了CodeForces 346C题目的解决方案，题目要求将数字a通过特定操作转化为b，探讨了贪心算法的应用及优化策略，详细解释了如何高效地减少操作次数。

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题目链接：http://codeforces.com/contest/347/problem/E
You are given a sequence of positive integers x1, x2, ..., xn and two non-negative integers a and b. Your task is to transform a into b. To do that, you can perform the following moves:

subtract 1 from the current a;
subtract a mod xi (1 ≤ i ≤ n) from the current a.
Operation a mod xi means taking the remainder after division of number a by number xi.

Now you want to know the minimum number of moves needed to transform a into b.

Input
The first line contains a single integer n (1 ≤ n ≤ 105). The second line contains n space-separated integers x1, x2, ..., xn(2 ≤ xi ≤ 109). The third line contains two integers a and b (0 ≤ b ≤ a ≤ 109, a - b ≤ 106).

Output
Print a single integer — the required minimum number of moves needed to transform number a into number b.

Examples
input
3
3 4 5
30 17
output
6
input
3
5 6 7
1000 200
output
206

/*
CodeForces346  C. Number Transformation II
给你两个数a,b
1.每次对于当前的a减去1
2.每次对于当前的a减去 a%(ta[i])
求最少多少次能得到b
类似于贪心,每次减去尽可能多的值。但是一直TLE- -.   后来可以发现a~a-a%ta[i]的所有值
如果减去a%ta[i],都会等于同一个值。  如果当a-a%ta[i] < b时,ta[i]可以说在后面的
搜索就没有作用了.于是乎把ta[i]除去.
hhh-2016-04-16 17:15:20
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <algorithm>
#include <functional>
#include <math.h>
using namespace std;
#define lson  (i<<1)
#define rson  ((i<<1)|1)
typedef long long ll;
const int mod = 1000000009;
const int maxn = 100040;
int ta[maxn];
int main()
{
    int n;
    while( scanf("%d",&n) != EOF)
    {
        int num = 0;
        int a,b;
        for(int i =0; i < n; i++)
        {
            scanf("%d",&ta[i]);
        }
        sort(ta,ta+n);
        n = unique(ta,ta+n)-ta;
        scanf("%d%d",&a,&b);
        while(a > b)
        {
            int cur = a - 1;
            for(int i =n-1; i >= 0; i--)
            {
                if(a-a%ta[i] >= b)
                    cur = min(a-a%ta[i],cur);
            }
            a = cur;
            num ++;
            if(a == b) break;
            while(n)
            {
                if(a-a%ta[n-1] < b)
                    n--;
                else
                    break;
            }
        }
        printf("%d\n",num);
    }
    return 0;
}