HDU5943-Kingdom of Obsession

Kingdom of Obsession

                                                                          Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                                     Total Submission(s): 1162    Accepted Submission(s): 356

Problem Description

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

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

xmody=0

Is there any way to satisfy everyone's requirement?

 

Input

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

Every test case contains one line with two integers n, s.

Limits
1≤T≤100.
1≤n≤109.
0≤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 y equals 'No'.

 

Sample Input

25 144 11

 

Sample Output

Case #1: NoCase #2: Yes

 

Source

2016年中国大学生程序设计竞赛（杭州）

 

题意：有n个人，每个人的标号以此是s+1到s+n，要求将所有人重新排序之后满足每个人的位置y能够保证被他的标号整除，就是数组重排之后满足每一位的a[i]%i==0

解题思路：一定区间之内最多只有一个质数，所以多于一个质数的直接NO（差不多任意两个相邻之间的素数的差不会超过500），其他情况可以直接二分图匹配去跑，把这段区间和1~n作为2*n个节点直接n^2建图跑匈牙利就行（这段区间和1~n可能有重叠的部分，这段是要删去的，重叠部分一定满足所以不需要匹配，所以最后图中的节点就是一边时1到min(n,s)一边是max(n,s)到n+s）

#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 longconst 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>550)        {            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;}