K - Kingdom of Obsession

原创于 2019-10-30 18:40:08 发布 · 289 阅读

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探讨一个关于整数排列的数学谜题，在一个由连续整数组成的王国中，每个成员希望根据特定条件占据特定位置，算法挑战在于判断是否能实现所有人的愿望。

There is a kindom of obsession, so people in this kingdom do things very strictly.

They name themselves in integer, and there are nn people with their id continuous (s+1,s+2,⋯,s+n)(s+1,s+2,⋯,s+n) standing in a line in arbitrary order, be more obsessively, people with id xx wants to stand at ythyth position which satisfy

xmody=0xmody=0

Is there any way to satisfy everyone's requirement?

Input

First line contains an integer TT, which indicates the number of test cases.

Every test case contains one line with two integers nn, ss.

Limits
1≤T≤1001≤T≤100.
1≤n≤1091≤n≤109.
0≤s≤1090≤s≤109.

Output

For every test case, you should output 'Case #x: y', where x indicates the case number and counts from 1 and y is the result string.

If there is any way to satisfy everyone's requirement, y equals 'Yes', otherwise yequals 'No'.

Sample Input

2
5 14
4 11

Sample Output

Case #1: No
Case #2: Yes

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
 
using namespace std;
 
#define LL long long
const int INF=0x3f3f3f3f;
 
int n,ss;
int x[559],y[559];
int s[559],nt[260009],e[260009];
int visit[559];
bool path(int k)
{
    for(int i=s[k]; ~i; i=nt[i])
    {
        int ee=e[i];
        if(!visit[ee])
        {
            visit[ee]=1;
            if(y[ee]==-1||path(y[ee]))
            {
                y[ee]=k;
                x[k]=ee;
                return 1;
            }
        }
    }
    return 0;
}
 
void MaxMatch()
{
    int ans=0;
    memset(x,-1,sizeof(x));
    memset(y,-1,sizeof(y));
    for(int i=1;i<=n;i++)
    {
        if(x[i]==-1)
        {
            memset(visit,0,sizeof(visit));
            if(path(i)) 
			ans++;
        }
    }
    if(ans==n) 
	printf("Yes\n");
    else 
	printf("No\n");
}
 
int main()
{
    int t,cas=0;
    scanf("%d",&t);
    while(t--)
    {
        printf("Case #%d: ",++cas);
        scanf("%d%d",&n,&ss);
        if(ss<=1)
        {
            printf("Yes\n");
            continue;
        }
        if(ss<n) 
		swap(ss,n);
        if(n>100)
        {
            printf("No\n");
            continue;
        }
        int cnt=1;
        memset(s,-1,sizeof(s));
        for(int i=ss+1;i<=ss+n;i++)
        { 
            for(int j=1;j<=n;j++)
            {
            	if(i%j==0) 
				{
					nt[cnt]=s[i-ss];
					s[i-ss]=cnt;
					e[cnt++]=j;
				}
            }
		} 
        MaxMatch();
    }
    return 0;
}