CodeForces The Endless River

本文介绍了一个有趣的编程竞赛问题——河流竞赛。题目设定在一个圆形的‘无尽之河’上,两位参与者以不同的速度前进并放置标记。任务是计算两者放置标记在同一位置所需的最短时间。文章提供了完整的代码实现及样例解释。

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The Endless River
Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
 

Description

standard input/output
Statements

You've possibly heard about 'The Endless River'. However, if not, we are introducing it to you. The Endless River is a river in Cambridge on which David and Roger used to sail. This river has the shape of a circular ring, so if you kept sailing forward you wouldn't find an end, and that's why it was called 'endless'. The river is exactly n meters long. Each minute, David moves d meters forward while Roger moves r meters forward; they start sailing from the same position at the same time, moving in the same direction. At the beginning of each minute, each of them leaves a marker at his current place (they don't put any markers at the beginning of the race). Determine the number of minutes before two markers (of different people) are placed at the same position.

Input

The input consists of several test cases. The first line of the input contains a single integer T, the number of the test cases. Each of the following lines represents a test case and contains three space-separated integers nd and r (1 ≤ n, d, r ≤ 100000) denoting the length 'in meters' of the river, David's speed and Roger's speed (as explained in the statement above) respectively.

Output

For each test case print a single line containing one integer: the number of minutes before two markers (of different people) are placed at the same position.

Sample Input

Input
4
8 2 3
8 3 2
5 1 4
100 1 1
Output
3
3
3
1

Hint

In the first case, we have n = 8, d = 2 and r = 3. They start at the first space which is denoted with 'start'. Red color denotes David's markers, green for Roger's markers, blue denotes that two markers (of different people) are placed. At the beginning (0 minutes passed), no markers are placed. After 1 minute passes, David is at the third space and Roger is at the fourth. After 2 minutes pass, David is at the fifth space and Roger is at the seventh. After 3 minutes pass, David is at the seventh space and Roger is at the minute, so when David places his marker, two markers are placed at the seventh space and the answer is 3 minutes.

 

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

typedef long long LL;

int a[100005];
int b[100005];

int main()
{
    //freopen("input.txt","r",stdin);
    int t;
    cin>>t;
    while(t--)
    {
        int n,d,r;
        cin>>n>>d>>r;
        int f=1;
        int s=1;
        int m=0;
        a[1]=1;
        int flag=0;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        while(1)
        {
            m++;
            f=m*d%n;
            if(b[f]==1)  {break;}
            a[f]=1;

            s=m*r%n;
            if(a[s]==1)   {break;}
            b[s]=1;
        }
        cout<<m<<endl;
    }
}

  

转载于:https://www.cnblogs.com/Hyouka/p/5719722.html

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